WlLjL'S 

COMMERCIAL 
ARITHMETIC 


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WILL'S 
COMMERCIAL  ARITHMETIC 


PRESENTING 

THE  BEST  USAGE  IN  MODERN 
BUSINESS  PRACTICE 


BT 

WILLIAM  R.  WILL 

n 

FOR  THIRTY-TWO  YEARS  PRINCIPAL  OP  THE  MATHEMATICAL  DEPARTMENT 
OF  SADLER'S  BRYANT  AND  STRATTON  BUSINESS  COLLEGE 
BALTIMORE;  AND  AUTHOR  OF  THE  SADLER-ROWE 
SERIES  OF  COMMERCIAL  ARITHMETICS 


THE  GREGG  PUBLISHING  COMPANY 

NEW  YORK  CHICAGO  SAN   FRANCISCO 


COPYRIGHT,     1913,    BY 
THE    GREGG    PUBLISHING    COMPANY 


f    -V 


THH-PLIMPTON-PRESS 
NORWOOD- MASS-U-S. A 


PREFACE 

IT  has  been  found  extremely  difficult,  if  not  impossible,  for 
the  average  student  of  arithmetic  to  remember  its  many  topical 
rules  and  explanations,  and  to  exercise  due  discrimination  in  giving 
to  each  its  local  application.  "Will's  Commercial  Arithmetic" 
has  therefore  ignored  this  unsatisfactory  method  of  treatment, 
and  has  substituted  the  few  cardinal  principles  which  underlie  all 
arithmetical  processes,  and  which  can  be  used  in  the  solution  and 
explanation  of  all  problems.  The  learner  is  thus  liberated  from 
slavish  adherence  to  mechanical  rules  and  hackneyed  explana- 
tions which  are  soon  forgotten,  and  is  properly  cultured  in  the 
exercise  of  his  reasoning  faculties  in  the  solution  of  problems,  the 
effect  of  which  is  permanent. 

The  only  conceivable  modifications  of  any  given  number  in  a 
problem  are  effected  either  by  increasing  it  or  by  decreasing  it. 
The  law  of  increase  or  decrease  is  uniform  in  its  application. 
Addition  and  multiplication  are  the  only  known  numerical  pro- 
cesses for  increasing  a  number,  the  former  being  employed  when 
the  increase  is  by  one  or  more  given  unequal  components,  and 
the  latter  when  by  equal  components.  Conversely,  subtraction 
and  division  are  the  only  known  means  for  diminishing  a  num- 
ber, the  former  being  employed  when  the  decrease  is  by  one  or 
more  given  unequal  components,  and  the  latter  when  by  equal 
components.  Each  of  the  four  preceding  processes  is  thus  seen 
to  be  invariably  limited  to  its  own  prerequisite  conditions ;  and 
the  proper  process  to  select  will  depend  upon  the  particular  pre- 
requisite conditions  which  are  given.  The  intelligent  solution 
of  a  problem  can  therefore  involve  nothing  further  than  the 
identification  of  its  numerical  terms,  a  knowledge  of  the  specific 
process  to  which  these  identified  terms  invariably  belong,  and 
the  mechanical  performance  of  that  process. 


314603 


iv  PREFACE 

The  above  method  of  treatment  of  all  problems,  though  a 
departure  from  the  traditional  standards,  is  not  an  untested 
theory,  but  a  thoroughly  demonstrated  success.  It  is  the  ripe 
fruitage  of  forty-eight  years'  experience  in  the  class-room,  during 
the  last  thirty-two  of  which  the  author  has  been  engaged  in 
teaching  commercial  arithmetic  exclusively  in  one  of  the  oldest 
and  most  reputable  business  colleges  in  the  United  States. 

To  his  fellow  teachers  who  have  been  disappointed  either  in 
the  lack  of  permanency  or  of  quality  in  the  results  hitherto  accom- 
plished by  the  conventional  treatment  of  this  most  important 
study,  the  methods  of  this  volume  are  confidently  submitted  as  an 
efficient  means  for  the  accomplishment  of  entirely  satisfactory 
results  in  the  future. 

THE  AUTHOR 


CONTENTS 


INTRODUCTION  PAGE 

Preliminary  Definitions 1 

NUMERICAL  EXPRESSION 

Reading  Numbers 2 

Writing  Numbers   4 

Roman  Numerals   6 

NUMERICAL  OPERATIONS 

Addition 7 

Subtraction    9 

Multiplication    11 

Division    17 

United  States  Money 24 

REVIEW 

Relation  of  Numbers    29 

Review  Examples 34 

PROPERTIES  OF  NUMBERS 

Factoring    39 

Least  Common  Dividends  ....  41 

Greatest  Common  Divisors  ...  43 

Cancelation    45 

COMMON   FRACTIONS 

Introduction 46 

Definitions 47 

General  Principles 48 

Reductions 49 

Addition 55 

Subtraction    57 

Multiplication 60 

Division    63 

Relation  of  Fractions   69 

Fractional  Comparisons 71 

Review  Examples 75 


DECIMAL  FRACTIONS  PAGE 

Introduction 77 

Definitions 78 

General  Principles 79 

Reading  Decimals 81 

Writing  Decimals   81 

Reductions 82 

Addition 85 

Subtraction    87 

Multiplication 88 

Division    89 

Review  Examples 92 

COMPOUND  NUMBERS 

Introduction 94 

Definitions 95 

Tables  of  Measures 95 

Reductions 

Denominate  Integers   100 

Denominate  Fractions 102 

Addition 104 

Subtraction    105 

Multiplication 107 

Division   108 

Review  Examples 110 

Metric  System Ill 

MEASUREMENTS 

General  Surfaces  116 

Carpeting,  Paving,  Roofing  ...  118 

Boards,  Joists,  etc 121 

Volumes,  generally 122 

Bins,  Cisterns,  Wells    125 

PERCENTAGE 

Introduction 129 

Definitions. .  129 


VI 


CONTENTS 


PERCENTAGE  (Continued]  PAGE 

Identification  of  Terms    130 

Relation  of  Terms 131 

To  find  Percentages   132 

To  find  Rates 135 

To  find  Bases 137 

Review  Examples 145 

APPLICATIONS   OF  PERCENT- 
AGE 

Profit  and  Loss   148 

Trade  Discount 154 

Ordinary  Interest   161 

Percentage  Method 162 

Aliquot  Method 165 

Accurate  Interest   174 

Relation  of  Terms 175 

Compound  Interest 181 

True  Discount 185 

Bank  Discount    189 

To  find  Proceeds 190 

To  find  Face  of  Note 193 

Partial  Payments 

United  States  Rule 195 

Merchants'  Rule  197 

Commission 200 

Stocks  and  Bonds 205 

Stock  Investments    .  211 


APPLICATIONS  OF   PERCENT- 
AGE (Continued)  PAQE 

Exchange    214 

Domestic    216 

Foreign  218 

Bankruptcy    221 

Insurance 223 

Taxes 224 

Duties  or  Customs 225 

MISCELLANEOUS 
Proportion 

Introduction 227 

Simple  Proportion 228 

Compound  Proportion 231 

Equation  of  Payments 233 

Averaging  Accounts 241 

Averaging  Accounts  Sales 247 

Cash  Balance 248 

Partnership    252 

APPENDIX 

Aliquot  Calculations 267 

Addition  by  Grouping 272 

Cross  Multiplication 274 

Powers  and  Roots 274 

Answers  . .  283 


COMMEECIAL    ARITHMETIC 


INTRODUCTION 

1.  Arithmetic  is  that  branch  of  knowledge  which  treats  of 
numbers.      It  explains  the  possible  relations  between  the  numer- 
ical expression  of  one  given  quantity  and  that  of  another  given 
quantity,  and  how  those  relations  may  be  utilized  to  find  the 
numerical  expression  of  a  required  quantity. 

2.  Quantity  is  that  property  of  matter  which  may  be  meas- 
ured, weighed,  valued  or  counted,  and  which  may  be  expressed 
by  a  number. 

3.  A  number  is  the  expression  of  a  quantity  by  the  use  of 
figures. 

NOTE  1.  (a)  A  number  is  said  to  be  abstract  when  the  kind  of 
quantity  is  not  expressed,  but  only  its  extent,  as  three,  five  times;  (6)  to  be 
concrete  when  the  kind  of  quantity  is  expressed  as  well  as  its  extent,  as  five 
gallons,  seven  yards;  (c)  to  be  simple  when  a  concrete  quantity  is  expressed 
in  only  one  denomination,  as  three  bushels,  twelve  barrels;  (d)  to  be  com- 
pound when  one  concrete  quantity  is  expressed  by  the  use  of  two  or  more 
different  denominations,  as  three  pounds  and  nine  ounces  of  sugar,  six  gallons 
three  quarts  and  one  pint  of  water;  (e)  to  be  integral  when  a  quantity  is 
expressed  in  one  or  more  whole  units,  as  six,  nine  pounds;  (/)  to  be 
fractional  when  a  quantity  is  expressed  in  one  or  more  parts  of  a  whole 
unit,  as  one-half,  three-fourths. 

NOTE  2.  When  one  number  is  placed  in  contrast  with  one  or  more  other 
numbers  it  is  said  to  be  (a)  like  when  it  expresses  a  quantity  of  the  same 
name,  as  five  yards  is  a  like  number  to  eight  yards;  (b)  and  it  is  said  to 
be  unlike  when  it  expresses  a  quantity  of  a  different  name,  as  eight  inches 
is  an  unlike  number  to  five  gallons. 

4.  A    unit   is  one  thing  of    the   quantity    expressed    by   a 
number.     Thus,  the  unit  of  eight  yards  is  one  yard,  of  twelve 
miles  is  one  mile. 

5.  A    problem    in    arithmetic    is    a    statement     containing 
numerical  terms  which  are  mutually  related,  and  which  requires 


READING1.  NUMBERS 

that  relation  to  be  perfected  by  finding  an  omitted  term,  called 
the  answer. 

NOTE.  The  separate  unequal  components  and  their  total  are  distinctive 
terms  of  addition  (if  a  total  is  required),  or  of  subtraction  (if  a  total  is  given). 
One  equal  component,  the  number  of  equal  components,  and  the  total  of 
equal  components,  are  distinctive  terms  of  multiplication  (if  a  total  is 
required),  or  of  division  (if  a  total  is  given).  These  five  terms  are  the 
alphabet  of  arithmetic. 

6.  The  analysis  of  a  problem  is  the  process  of  separating 
it   into   its  given  numerical  terms,  ascertaining  their  character, 
discovering  their  relation  to  each  other,  and  thus  deducing  the 
appropriate  operation  to  find  the  required  term  which  will  com- 
plete that  relation. 

NOTE.  A  completed  relationship  of  arithmetical  terms  is  a  numerical 
total  and  the  several  equal  or  unequal  numerical  terms  which  constitute 
that  total. 

7.  The   solution   of  a   problem   is    the    process    of    finding 
the  required  term  which  will   complete  the  relationship  which 
exists  between  the  given  terms. 

NOTE.  A  relation  is  completed  by  finding  a  required  component 
which  will  complete  a  given  total,  by  finding  a  required  total  which  will 
include  all  of  its  given  components,  or  by  finding  the  number  of  given  equal 
components,  which  are  contained  in  a  given  total. 

READING  NUMBERS 

8.  Reading    numbers  is  the  mental  conception  or  the  vocal 
expression  of  the  value  of  a  written  number. 

9.  Numbers  are  commonly  expressed  by  the  use  of  the  ten 
following  figures,  used  singly  or  in  combination: 

0  123456789 

Naught       One       Two       Three       Four       Five       Six       Seven       Eight       Nine 

The  first  of  the  above  figures  is  called  naught  or  cipher,  and 
it  expresses  the  absence  of  value.  The  remaining  nine  figures 
are  called  significant  figures,  (a)  When  written  by  itself  or  at 
the  extreme  right  of  a  number,  each  figure  expresses  the  value 
placed  underneath  it  in  the  above  illustration.  Numbers 
greater  than  nine  are  expressed  by  a  systematic  combination 


READING    NUMBERS  3 

of  the  above  figures.  Thus,  (6)  to  increase  any  of  the  above 
figures  tenfold,  place  one  figure  at  the  right  of  it;  (c)  to  increase 
it  a  hundredfold,  place  two  figures  at  the  right  of  it;  (d)  to  in- 
crease it  a  thousandfold,  place  three  figures  at  the  right  of  it;  etc. 
(e)  The  local  value  of  a  figure  is  determined  by  its  position,  in- 
creasing tenfold  for  each  place  that  it  is  moved  to  the  left; 
(/)  and  decreasing  tenfold  for  each  place  that  it  is  moved  to 
the  right,  (g)  The  position  which  a  figure  occupies  in  a  number, 
counting  from  the  right,  determines  its  value  and  is  called  its 
order. 

ILLUSTRATIVE  EXERCISE 

10.  Read  4976. 

EXPLANATION.  Commencing  with  the  left-hand  figure  (4)  read  it  and 
name  its  local  value  (4  thousand),  as  it  has  three  figures  at  its  right  [9,  d]; 
then  read  the  next  following  figure  and  name  its  local  value  (9  hundred),  as 
it  has  two  figures  at  its  right  [9,  c];  then  read  the  next  following  figure  and 
name  its  local  value  (7ty  or  7  tens,  ty  being  a  corruption  of  ten),  as  it  has  one 
figure  at  its  right  [9,  6];  then  read  the  next  following  and  last  figure  (6), 
omitting  its  name  (units),  as  it  is  understood  [9,  a];  thus  making  the  entire 
expression  read  "  four  thousand  nine  hundred  seventy-six." 

11.  Numbers  composed  of  more  than  three  figures  are  read  in 
sets  of  three  figures  each,     (a)  The  first  set  of  three  figures,  count- 
ing from  the  right,  is  called  the  Units1  Period;  (b)  the  second  set 
of  three  figures  from  the  right  is  called  the  Thousands'  Period; 
(c)  the  third  set  of  three  figures,  the  Millions'  Period;    (d)  the 
fourth  set,  the  Billions'  Period;    (e)  the  fifth  set,  the  Trillions' 
Period;  (/)  the  sixth,  the  Quadrillions'  Period;  etc. 

SUGGESTION.  To  read  and  write  numbers  with  facility,  it  is  necessary 
that  the  above  periods  be  memorized  until,  without  hesitation,  they  can  be 
repeated  in  regular  order,  forward  from  the  highest  period,  or  backward  from 
the  lowest. 

EXERCISES 
Express  the  following  numbers  in  written  or  spoken  words : 


1.  9 

6.  72 

9.  426 

13.  7836 

17.  8003 

2.  14 

6.  45 

10.  865 

14.  5492 

18.  4600 

3.  18 

7.  97 

11.  309 

16.  6285 

19.  3280 

4.  36 

8.  58 

12.  200 

16.  9058 

20.  7025 

4  WRITING    NUMBERS 

ILLUSTRATIVE  EXERCISE 

12.  Read  26£63728£9Q219. 

EXPLANATION.  To  discover  the  name  of  each  period  in  the  given  number 
and  the  figures  which  compose  it,  commence  at  the  right  and  separate  it  into 
periods  of  three  figures  each,  as  follows:  26,463,728,496,219.  As  the  right- 
hand  period  (219)  is  pointed  off,  think  of  its  name  (units);  as  the  second 
period  (496)  is  pointed  off,  think  of  its  name  (thousands) ;  as  the  third  period 
(728)  is  pointed  off,  think  of  its  name  (millions),  and  so  continue  until  the 
left-hand  period  (26)  is  reached,  which  is  thus  found  to  be  trillions. 

Then  read  the  number  expressed  by  the  left-hand  period  as  if  it  stood 
alone,  calling  its  name  (twenty-six  trillion) ;  then  read  the  number  expressed 
by  the  next  following  period  as  if  it  stood  alone,  calling  its  name  (four  hundred 
sixty-three  billion) ;  then,  similarly,  the  next  following  period  (seven  hundred 
twenty-eight  million);  then  the  next  period  (four  hundred  ninety-six  thou- 
sand); and  finally  the  last  period,  omitting  its  name  (units)  as  it  is  under- 
stood (two  hundred  nineteen). 

EXERCISES 

Express  the  following  numbers  in  written  or  spoken  words: 

1.  2785  11.  678349526  21.  51836274813429 

2.  16298  12.  2817462785  22.  200070600083294 

3.  34276  13.  7132975138  23.  56000000400000 

4.  416298  14.  231678261582  24.  290867320000078 
6.  728341  16.  57281334967864  25.  689712000091863 

6.  2685913  16.  1963475162834748  26.  500030007100006 

7.  5873265  17.  5600328176003452  27.  8620053000007008 

8.  37264897  18.  92086005027316814  28.  65000000002009 

9.  52819624  19.  51378260000009162  29.  1834000000007 10006 
10.  83429857  20.  60000520080076003  30.  2000009000070005 

WRITING  NUMBERS 

13.  (a)  Writing  numbers  is  the  expression  in  figures  of  the 
value  of  a  number  when  given  in  words.     (6)  Knowing  the  names 
of  the  several  periods  and  their  relative  positions  [11],  any  num- 
ber is  written  by  commencing  with  its  highest  period  and  writing 
it  as  if  it  stood  alone,  then  writing  its  next  highest  period  as  if  it 
stood  alone,  and  so  continuing  with  period  after  period  in  regular 
order  until  the  lowest  period  is  written. 

NOTE,  (a)  If,  after  writing  the  highest  period,  any  subsequent  period 
should  be  found  to  contain  fewer  than  three  figures,  supply  the  deficiency  by 


WRITING   NUMBERS  5 

prefixing  the  requisite  number  of  ciphers.     (6)  If  an  entire  period  is  omitted, 
supply  three  ciphers  for  its  three  vacant  orders. 

ILLUSTRATIVE  EXERCISE 

Express  in  figures,  twenty-seven  quadrillion,  nine  hundred 
sixty-eight  trillion,  seventy-two  million,  eight  thousand  forty. 

EXPLANATION.  Commencing  with  the  highest  period  of  the  given  num- 
ber (quadrillions),  write  it  as  if  it  stood  alone  (27  .  .  .);  then  write  the  next 
lower  period  (trillions)  as  if  it  stood  alone  (27,968  .  .  .)  then  write  the  next 
lower  period  (billions)  and  since  no  billions  are  expressed  in  the  given  number 
[Note,  6,13],  supply  three  ciphers  for  its  three  vacant  orders  (27,968,000  .  .  .); 
then  write  the  next  lower  period  (millions),  and  as  seventy-two  requires  only 
two  figures  to  express  it  and  each  period  except  the  highest  must  contain  three 
figures  [Note,  a,  13],  supply  the  deficiency  by  prefixing  one  cipher  (27,968,000,- 
072  .  .  .);  then  write  the  next  lower  period  (thousands),  and  as  eight  requires 
only  one  figure  to  express  it,  and  each  period  except  the  highest  must  contain 
three  figures,  supply  the  deficiency  by  prefixing  two  ciphers  (27,968,000,072,- 
008  .  .  .);  then  write  the  next  lower  period  (units),  and  as  forty  requires 
only  two  figures  to  express  it,  prefix  one  cipher,  obtaining  27,968,000,072,- 
008,040  as  the  complete  numerical  expression. 

EXERCISES 

Express  in  figures  the  following  numbers: 

1.  Six  thousand,  three  hundred  fifty-two. 

2.  Forty-nine  thousand,  eight  hundred  seventy-six. 

3.  Two  hundred  nine  thousand,  seven  hundred. 

4.  Five  hundred  eight  thousand,  sixty-three. 
6.  Ninety  thousand,  eight. 

6.  Eighty-three  thousand,  six  hundred  ninety-five. 

7.  Twelve  million,  seventy-one  thousand  six. 

8.  Three  million,  sixteen  thousand,  two  hundred  eight. 

9.  Nineteen  million,  three  hundred  fifty-eight  thousand 
forty. 

10.  Twenty-five  million,  five  hundred  twenty-seven. 

11.  Four  trillion,  seventy-six  million,  two  hundred  fifty- 
three  thousand,  seven. 

12.  Fifteen  quadrillion,  two  hundred  ninety-one  trillion,  five 
billion,  seven-two  thousand,  sixteen. 


6  ROMAN    NUMERALS 

Express  in  figures  the  following  numbers: 

13.  Ninety-seven   quadrillion,    one   hundred    twenty-eight 
million,  four  hundred  thousand,  ninety-two. 

14.  Six  quadrillion,  seven  hundred  trillion,  eighty  billion, 
nine  hundred  fifty  thousand,  twenty. 

ROMAN  METHOD  OF  EXPRESSING  NUMBERS 

14.   The  Roman  method  of  expressing  numbers  is  dependent 
upon  the  following  seven  capital  letters,  used  as  numerals. 


I 

One 


V 

Five 


X 

Ten 


L 

Fifty 


c 

One  hundred 


D 

Five  hundred 


M 

One  thousand 


NOTE  1.  Placed  by  itself,  each  of  the  above  Roman  numerals  expresses 
the  value  written  underneath  it. 

NOTE  2.  Successive  repetitions  of  the  above  numerals  denote  repetitions 
of  the  value  which  they  express.  Thus,  II  express  two  ones,  or  two;  III, 
three  ones,  or  three;  XXX,  three  tens,  or  thirty;  CC,  two  hundred;  etc. 

NOTE  3.  The  combined  value  when  one  numeral  is  placed  at  the  left  of 
another  of  greater  value  expresses  the  difference  between  their  respective 
values.  Thus,  IV  expresses  the  difference  between  one  and  five,  or  four; 
IX  expresses  nine;  XL,  forty;  XC,  ninety;  etc. 

NOTE  4.  The  combined  value  when  one  numeral  is  placed  at  the  right  of 
another  of  greater  value  expresses  the  sum  of  their  respective  values.  Thus, 
VIII  expresses  eight;  XXV,  twenty-five;  LXII,  sixty-two;  etc. 

15.  The  manner  in  which  Roman  numerals  are  combined  to 
express  numbers  is  shown  in  the  following 

TABLE  OF  ROMAN   NUMERALS 


I 

One 

XIV 

Fourteen 

XC 

Ninety 

II 

Two 

XV 

Fifteen 

C 

One  hundred 

III 

Three 

XVI 

Sixteen 

CC 

Two  hundred 

IV 

Four 

XVII 

Seventeen 

ccc 

Three  hundred 

V 

Five 

XVIII 

Eighteen 

cccc 

Four  hundred 

VI 

Six 

XIX 

Nineteen 

D 

Five  hundred 

VII 

Seven 

XX 

Twenty 

DC 

Six  hundred 

VIII 

Eight 

XXX 

Thirty 

DCC 

Seven  hundred 

IX 

Nine 

XL 

Forty 

DCCC 

Eight  hundred 

X 

Ten 

L 

Fifty 

DCCCC 

Nine  hundred 

XI 

Eleven 

LX 

Sixty 

M 

One  thousand 

XII 

Twelve 

LXX 

Seventy 

MM 

Two  thousand 

XIII 

Thirteen 

LXXX 

Eighty 

MMM 

Three  thousand 

ADDITION  7 

NOTE  1.  No  numeral  should  be  repeated  more  than  three  successive 
times,  except  C  which  may  be  repeated  four  times,  but  no  more. 

NOTE  2.  Only  one  numeral  of  less  value  can  be  written  at  the  left  of 
another  of  greater  value. 

NOTE  3.  A  bar  placed  over  a  Roman  numeral,  or  a  combination  of  such 
numerals,  increases  its  value  a  thousandfold.  Thus  V  expresses  5000;  XXIV, 
24000;  CCXLV1DXII,  246512. 

NOTE  4.  Vacant  orders  are  not  expressed  as  is  found  necessary  when 
figures  are  employed.  There  is  no  Roman  numeral  to  express  naught. 

ILLUSTRATIVE  EXERCISE 
16.   Express  3675  by  the  use  of  Roman  numerals. 

EXPLANATION.  First  write  the  highest  order,  three  thousand  (MMM); 
then  the  next  lower  order,  six  hundred  (MMMDC);  then  the  next  lower 
order,  seven  tens  or  seventy  (MMMDCLXX);  and  finally  the  lowest  order, 
five  units  (MMMDCLXXV). 

Express  the  following  by  the  use  of  Roman  numerals: 

1.  12  6.  92  9.  219  13.  729  17.  1468 

2.  27  6.  69  10.  188  14.  942  18.  2372 

3.  16  7.  35  11.  347  15.  638  19.  1090 

4.  45  8.  74  12.  406  16.  826  20.  3007 

Express  the  following  by  the  use  of  figures : 

21.  LXXXVI  25.  DCCXXVI  29.  MCXLVIII 

22.  CCXLIV  26.  MCCLVII  30.  MDCCCLV 

23.  DCLXIX  27.  DCCCCXXI  31.  MDCCCCIX 

24.  CCCCXII  28.  MMCXXXV  32.  DCCCLXXXIX 


ADDITION 

17.   Addition  is  the  name  of  the  process  for  finding  the  total 
of  two  or  more  unequal  quantities. 

NOTE  1.     The  combined  value  of  all  the  quantities  to  be  added  is  called 
the  sum,  or  total,  or  amount. 

NOTE  2.     (a)  The  sign  of  addition  is  an  erect  cross  (+)  and  is  called  plus. 

(b)  When  placed  between  quantities,  it  denotes  that  they  are  to  be  added. 

(c)  The  sign  of  equality  is  two  short  horizontal  lines  ( =  ),  and  is  read  equals  or 
equal,     (d)  The  sign  of  equality  denotes  that  the  combined  expression  on 
the  left  of  it  is  equal  to  the  combined  expression  on  the  right  of  it. 


8  ADDITION 

18.  Principles  of  Addition.     1.  Only  like  numbers  [Note  2,  a,  3], 
or  like  orders  [9,  g],  can  be  added.     2.  The  sum  expresses  a  quantity 
of  the  same  name  as  the  quantities  added. 

ILLUSTRATIVE  EXAMPLE 

19.  Add  8167,  59,  374,  2742,  19,  3816,  7  and  34. 

SOLUTION  EXPLANATION.     So  place  the  numbers  to  be  added  that 

figures  of  the  same  order  [9,  g]  shall  fall  in  the  same  column 
81o7  [Prin.  1,  18],  and  draw  a  horizontal  line  beneath.     Beginning 

59  with  the  right-hand  or  units'  column,  find  the  sum  of  4  +  7 

374  +6  +  9+2+4  +  9  +  7,  amounting  to  48  units,  or  4  tens 

2742  anc*  **  units.     Place  the  8  units  beneath  the  horizontal  line 

1  Q  and  under  the  units'  column,  and  carry  the  4  tens  to  the  tens' 

column,  that  is,  mentally  add  the  4  carried  tens  with  the 
oolt  figures  in  the  tens'  column.     Thus,  4   (the  carried  figure) 

7  +  34-i-j-i+4_f_7  +  5_|_6=31    tens,    or    3    hundred 

34  and  1  ten.     Place  the  1  ten  beneath  the  line  and  under  the 

tens>   c°lumn>   and  carry  the  3  hundred  to  the  hundreds' 
234  column,  as  follows:  3  (the  carried  figure)  +8  +  7  +  3  +  1  = 

22  hundred,  or  2  thousand  and  2  hundred.  Place  the  2  hun- 
dred beneath  the  line  and  under  the  hundreds'  column,  and  carry  2  thousand 
to  the  thousands'  column,  as  follows:  2  (the  carried  figure)  +3  +  2  +  8  =  15 
thousand.  As  this  is  the  last  column  to  be  added,  place  the  entire  result 
beneath  the  line  in  such  a  manner  that  the  5  thousand  shall  fall  under  the 
thousands'  column. 

NOTE  1.  To  prove  addition,  add  each  column  in  the  opposite  direction 
to  that  first  employed.  If  the  same  results  are  obtained,  the  answer  is  pre- 
sumed to  be  correct. 

NOTE  2.  When  adding,  write  the  carrying  figure  under  the  column  from 
which  it  was  obtained,  as  shown  in  the  illustrative  example,  that  it  may  also 
be  included  in  the  proof. 

NOTE  3.  In  adding  columns,  acquire  the  habit  of  pronouncing  results 
without  naming  the  figures  from  which  they  have  been  obtained.  Thus,  in 
adding  the  units'  column  of  the  illustrative  solution,  it  is  better  to  read  4,  11, 
17,  26,  28,  32,  41,  48,  than  to  spell  4  and  7  are  11,  11  and  6  are  17,  etc. 

EXAMPLES  FOR  PRACTISE 

Find  the  sum  of 

1.  58,  36,  75,  92,  85,  36,  27,  48,  62,  59,  23,  37,  56,  18 

2.  326,  62,  927,  35,  5268,  76,  248,  9,  628,  3253,  769 

3.  8296,  573,  2894,  37,  518,  6275,  93,  235,  58,  617,  75 


SUBTRACTION  9 

Find  the  sum  of 

4.  5368,  7864,  3295,  48,  2834,  376,  58,  675,  49,  3278 

5.  18237,  625,  34296,  5867,  928,  2678,  18326,  754 

6.  Three   thousand   two    hundred   ninety-five;    seven  hun- 
dred sixteen;    one  thousand  forty-eight;    thirty-five  thousand 
ninety-seven;    eight  thousand  three  hundred  five;    sixty-nine 
thousand  eight;  eighty  thousand  four;  forty  thousand  one  hun- 
dred;  two  hundred  six  thousand  seven  hundred  three;    four 
hundred  thousand  five  hundred  twenty-eight. 

20.  It  is  frequently  found  necessary  to  add  numbers  which  are 
written  horizontally,  or  in  some  other  way  than  in  the  usual  ver- 
tical arrangement.     To  find  the  total  of  numbers  thus  irregularly 
arranged,  first  add  all  the  right-hand  orders;    next  add  all  the 
second  orders  from  the  right;  next,  all  the  third  orders  from  the 
right,  etc.;  setting  down  and  carrying  in  the  usual  manner. 

EXAMPLES  FOR  PRACTISE 

Find  the  total  by  adding  horizontally : 

1.  562,  375,  812,  492,  716       5.   826,  9187,  46,  3285,  768 

2.  283,  826,  749,  514,  328       6.   96,  487,  3256,  7834,  279 

3.  617,  482,  627,  835,  569       7.   38,  46,  298,  84,  2637,  526 

4.  986,  789,  582,  796,  328       8.   3682,  597,  2876,  736,  98 

SUBTRACTION 

21.  Subtraction  is  the  name  of    the    process  for  finding  the 
difference  between  two  unequal  quantities. 

NOTE  1.  (a)  Subtraction  involves  the  reverse  use  of  the  same  terms  as 
have  been  already  considered  in  addition  [Note  1,  17].  The  names  of  these 
terms  of  addition  are  changed,  however,  to  accord  with  the  reverse  process 
employed  in  their  use.  (6)  The  total  value  of  all  the  unequal  components 
formerly  called  the  sum,  is  now  distinguished  as  the  minuend,  which  means 
the  quantity  to  be  diminished;  (c)  the  given  component  (or  the  sum  of  the 
given  components  if  more  than  one  is  given)  is  now  called  the  subtrahend,  which 
means  quantity  to  be  subtracted;  (d)  and  the  required  component  is  now 
called  the  remainder,  which  means  that  (quantity)  which  remains.  Hence, 

NOTE  2.  (a)  The  minuend  is  the  quantity  to  be  diminished;  (6)  the 
subtrahend  is  the  quantity  to  be  taken  from  the  minuend;  (c)  and  the  re- 
mainder or  difference  is  what  remains  of  the  minuend  after  the  subtrahend 
has  been  taken  from  it. 


10  SUBTRACTION 

NOTE  3.  (a)  The  sign  of  subtraction  is  a  short  horizontal  line,  as  follows 
(  — ).  (6)  It  is  called  minus;  (c)  it  means  less;  (d)  and  when  placed  between 
two  numbers,  it  indicates  that  the  number  at  its  right  is  to  be  subtracted  from 
the  number  at  its  left. 

22.  Principles   of  subtraction.     1.  The  difference  can  be  found 
only  between  like  numbers  [Note  2,  a,  3]  or  like  orders  [9,  g].     2.  The 
difference  between  like  numbers  or  like  orders  will  also  be  like.     3.  An 
increase  of  any  order  in  the  subtrahend  is  equivalent  to  a  correspond- 
ing decrease  of  the  similar  order  in  the  minuend.     4.    The  difference 
between  two  numbers  is  the  difference  of  the  several  orders  in  the 
subtrahend  separately  taken  from  their  respective  similar  orders  in  the 
minuend. 

ILLUSTRATIVE  EXAMPLE 

23.  Subtract  482  from  827. 

SOLUTION  EXPLANATION.  Write  the  less  number  under  the  greater 
BO  that  each  order  [9,  g]  of  the  less  number  shall  fall  under  a 
similar  order  of  the  greater.  Commencing  with  the  lowest 
order,  subtract  2  units  of  the  subtrahend  from  7  units  of  the 
minuend,  leaving  5  units  to  occupy  the  units'  order  of  the 
remainder. 

Then  subtract  the  next  higher  order  of  the  subtrahend  (8  tens)  from 
the  corresponding  next  higher  order  of  the  minuend  (2  tens),  and  since  the 
8  tens  of  the  subtrahend  cannot  be  taken  from  the  2  tens  in  the  minuend, 
"borrow"  1  hundred  from  the  8  hundred  in  the  minuend,  and  add  the  bor- 
rowed hundred  (or  10  tens)  to  the  2  tens,  thus  making  12  tens,  and  then 
subtract  8  tens  from  12  tens,  leaving  4  tens  to  occupy  the  tens'  order  of  the 
remainder. 

Lastly,  subtract  the  next  higher  order  of  the  subtrahend  (4  hundred) 
from  the  next  higher  order  of  the  minuend  (7  hundred),  that  is,  8  hundred 
diminished  by  the  previously  borrowed  1  hundred,  obtaining  3  hundred  as 
the  hundreds'  order  of  the  remainder. 

NOTE.  To  prove  subtraction,  add  the  remainder  (the  required  component, 
(Note  1,  d,  21)  to  the  subtrahend  (the  given  component,  Note  1,  c,  21).  The 
result  should  equal  the  minuend  (the  given  total,  Note  1,  6,  21). 

EXAMPLES  FOR  PRACTISE 
Find  the  difference  between 

1.  9426  and  4732  4.    79218  and  24835 

2.  8592  and  5278  5.    121829  and  72636 

3.  92816  and  37284  6.    187213  and  95847 


SUBTRACTION  11 

Find  the  difference  between 

7.  68329142  and  41768315  10.  918326781  and  40067007 

8.  50380674  and  27300407  11.  700400305  and  22080026 

9.  80000000  and  26040058  12.  602005009  and  41030070 

13.  Subtract  three  hundred  forty-five  thousand  seven  hun- 
dred nine  from  eight  hundred  four  thousand  six  hundred. 

14.  Subtract  one  million   eight   hundred   seven   thousand 
five  hundred  thirty-six  from  twenty-five  million  one  hundred 
thousand  ninety. 

15.  A  speculator  bought  a  tract  of  land  for  $52786  and 
afterwards  sold  it  for  $71000.     What  was  his  gain? 

16.  A  firm's  deposits  in  a  bank  amounted  to  $62309  and 
its  withdrawals  to  $18027.     How  much  did  the  firm  have  re- 
maining in  the  bank? 

17.  If  a  man  was  born  in  the  year  1843  and  died  in  the 
year  1911,  how  old  was  he  at  the  time  of  his  death? 

18.  A  house  was  sold  for  $28134  which  was  $2698  more 
than  the  sum  paid  for  it.     What  did  the  house  cost? 

24.  To  save  the  time  consumed  in  re-arranging  the  terms  of 
subtraction,  learners  should  acquire  the  art  of  finding  the  differ- 
ence between  two  numbers  in  whatever  position  they  may  have 
been  written.     This  may  be  done  as  follows :     Subtract  the  right- 
hand  order  of  the  less  number  from  the  right-hand  order  of  the 
greater ;  proceed  in  a  similar  manner  with  the  second  orders  from 
the  right;  with  the  third  orders  from  the  right,  etc.;  until  all  the 
orders   have  been  considered,  checking  off  each  order  as  it  is 
subtracted  to  facilitate  the  proper  identification  of  the  next  order 
to  be  considered. 

EXAMPLES  FOR  PRACTISE 

Subtracting  horizontally,  find  the  difference  between 

1.  92863  and  75392  4.  3289658  and  1832982 

2.  57029  and  20807  5.  8050027  and  4207009 

3.  80035  and  62008  6.   7200503  and  2830408 

MULTIPLICATION 

25.  Multiplication  is  the  name  of    the    process   for   finding 
the  total  of  two  or  more  equal  quantities. 


12 


MULTIPLICATION 


NOTE  1.  (a)  One  of  the  considered  equal  quantities  is  called  the  mul- 
tiplicand; (6)  the  number  of  equal  quantities  is  called  the  multiplier;  and 
(c)  the  total  of  all  the  considered  equal  quantities  is  called  the  product. 

NOTE  2.  The  multiplicand  and  multiplier  are  called  factors  (producers) 
of  the  product. 

NOTE  3.  (a)  The  sign  of  multiplication  is  an  oblique  cross  ( X ) .  (6)  It 
is  read  multiplied  by;  (c)  and  when  placed  between  numbers,  it  indicates 
that  they  are  to  be  multiplied. 

26.  Principles  of  multiplication.  1.  The  product  expresses  a 
quantity  of  the  same  kind  as  the  multiplicand  [Prin.  2,  18].  2. 
The  complete  product  equals  the  product  of  each  order  of  the  multi- 
plicand separately  multiplied  by  each  order  of  the  multiplier.  3. 
To  multiply  or  divide  the  multiplier  by  any  number,  the  multipli- 
cand remaining  unchanged,  is  equivalent  to  multiplying  or  dividing 
the  product  by  that  number.  4.  To  multiply  or  divide  the  multi- 
plicand by  any  number,  the  multiplier  remaining  unchanged,  is 
equivalent  to  multiplying  or  dividing  the  product  by  that  number. 


MULTIPLICATION  TABLE 


1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

1 

2 

4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

2 

3 

6 

9 

12 

15 

18 

21 

24 

27 

30 

33 

36 

3 

4 

8 

12 

16 

20 

24 

28 

32 

36 

40 

44 

48 

4 

5 

10 

15 

20 

25 

30 

35 

40 

45 

50 

55 

60 

5 

6 

12 

18 

24 

30 

36 

42 

48 

54 

60 

66 

72 

6 

7 

14 

21 

28 

35 

42 

49 

56 

63 

70 

77 

84 

7 

8 
9 

16 
18 

24 

27 

32 
36 

40 
45 
50 

48 

56 
63 

64 

72 

72 

80 

88 

96 

8 

54 

81 

90 

99 

108 

9 

10 

20 

30 

40 

60 

70 

80 

90 

100 

110 

120 

10 

11 

22 

33 

44 

55 

66 

77 

88 

99 

110 

121 

132 

11 

12 

24 

36 

48 

60 

72 

84 

96 

108 

120 

132 

144 

12 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

1 

MULTIPLICATION  13 

ILLUSTRATIVE  EXAMPLE 

27.  Multiply  976  by  8. 

SOLUTION          EXPLANATION.     Write  the  multiplier  (8)  under  the  lowest 
order  of  the  multiplicand  (6).     Commencing  with  this  lowest 
97 6         order  of  the  multiplicand,  separately  multiply  it  (6  units)  and 
8         each  successive  higher  order  (7  tens  and  9  hundred)  by  the 
multiplier  (8),  setting  down   and   carrying   as   usual.     Thus, 
8  times  6  units  equal  48  units,  or  4  tens  and  8  units.      Write 
the  8  units  beneath  the  line  and    under    the  units'   order  of  the  multi- 
plicand, and  carry  the  4  tens  to  8  times  7  tens  or  56  tens,  making  60  tens, 
or  6  hundred  and  0  tens.     Write  the  0  tens  beneath  the  line  and  under  the 
tens'  order  of  the  multiplicand,  and  carry  the  6  hundred  to  8  times  9  hundred 
or  72  hundred,  making  78  hundred.     As  this  completes  the  multiplication, 
write  this  final  result  in  full  beneath  the  line,  but  in  such  a  manner  that  the 
8  hundred  shall  fall  under  the  hundreds'  order  of  the  multiplicand. 

EXAMPLES  FOR  PRACTISE 
Multiply  Multiply 

1.  68734258  by  5  7.  84037028009  by  8 

2.  27893542  by  7  8.  70040006897  by  12 

3.  72486359  by  9  9.  40398002756  by  4 

4.  86957423  by  2  10.  920007100008  by  6 

5.  51627384  by  11  11.  53647298307  by  8 

6.  91827436  by  10  12.  80043090076  by  3 

ILLUSTRATIVE  EXAMPLE 

28.  Multiply  6723  by  894. 

SOLUTION  EXPLANATION.     First,    as   explained   in   27,    multiply 

the  successive  orders  of  the  multiplicand  (6723)  by  the 
units'  order  of  the  multiplier  (4)  as  if  it  stood  alone,  obtain- 
894  ing  26892  as  the  first  partial  product. 

26892  Next,  similarly  multiply  the  successive  orders  of  the 

fiQrQ7  multiplicand  (6723)  by  the  next  higher  order  of  the  mul- 

tiplier (9  tens)  as  if  it  stood  alone,  obtaining  60507  tens. 
53784  g0  piace  this  result  that  its  right-hand  order  (7  tens)  shall 

6010362  fall  under  the  tens'  order  of  the  preceding  partial  product 

[Prin.  1,  18]. 

Next,  similarly  multiply  the  successive  orders  of  the  multiplicand  (6723) 
by  the  hundreds'  order  of   the  multiplier  (8)  as  if  it  stood  alone,  obtain- 
ing 53784  hundred.     So  place  this  result  that  its  right-hand  order  (4  hundred) 
shall  fall  in  the  hundreds'  column  of  the  preceding  products.     [Prin.  1,  18.] 
The  sum  of  these  partial  products  (6010362)  will  be  the  complete  product. 


14  MULTIPLICATION 

NOTE  1.  The  first  figure  of  each  partial  product  when  properly  placed 
will  fall  directly  under  that  order  of  the  multiplier  which  was  used  to  produce 
it.  Thus,  in  the  above  solution,  the  figure  at  the  right  of  the  first  partial 
product  (26892),  which  is  2,  should  fall  under  that  order  of  the  multiplier  (4) 
which  produced  it.  The  figure  at  the  right  of  the  second  partial  product 
(60507),  which  is  7,  should  fall  under  that  order  of  the  multiplier  (9)  which  pro- 
duced it.  The  figure  at  the  right  of  the  third  partial  product  (53784),  which 
is  4,  should  fall  under  that  order  of  the  multiplier  (8)  which  produced  it. 

NOTE  2.  One  proof  of  multiplication  is  to  reverse  the  position  of  the 
multiplicand  and  multiplier,  and  multiply.  If  the  same  final  product  is 
obtained,  it  may  be  accepted  as  correct.  After  division  has  been  learned, 
another  method  of  proof  is  to  divide  the  obtained  product  by  either  of  its  fac- 
tors [Note  2,  25],  and  the  quotient  should  equal  its  other  factor. 

EXAMPLES  FOR  PRACTISE 

Multiply  Multiply 

1.  4768359  by  86  8.   683291482  by  8296 

2.  96378342  by  74  9.   503800209  by  6083 

3.  72839457  by  63  10.   700680005  by  4009 

4.  84976523  by  29  11.   829600706  by  9206 

5.  58297386  by  629  12.   920053008  by  70806 

6.  72682943  by  785  13.   438007056  by  40075 

7.  83795824  by  246  14.   320008062  by  32004 

15.  Find  the  total  weight  of  8965  barrels  of  flour  which  aver- 
age 196  pounds  per  barrel. 

16.  What  is  the  cost  of  building  374  miles  of  railroad  at  $34- 
768  per  mile? 

17.  A  man  earns  135  dollars  per  month  and  spends  46  dol- 
lars per  month.     How  much  does  he  save  in  9  months? 

18.  If  the  average  speed  of  a  steamer  is  365  miles  per  day, 
what  distance  can  it  traverse  in  137  days? 

29.  Continued  multiplication  is  the  successive  multiplication 
of  more  than  two  factors.  The  final  result  of  such  a  multiplica- 
tion is  called  a  continued  product. 

Thus,  if  816  be  multiplied  by  6,  and  the  resulting  product  by 
7,  the  process  is  called  a  continued  multiplication,  and  the  final 
result,  a  continued  product.  Continued  multiplication  is  em- 
ployed when  an  inconvenient  multiplier,  as  42,  can  be  changed  to 
two  or  more  convenient  multipliers,  the  product  of  which  is  equal 
to  42,  as  6  X  7. 


MULTIPLICATION  15 

ILLUSTRATIVE  EXAMPLE 

Multiply  986  by  63. 

SOLUTION  EXPLANATION.     One  of  the  factors  (63)  is  separable  into 

986          two  other  and  more  convenient  factors  (9  X  7).     Therefore, 
g          first  multiply  the  multiplicand   (986)  by  one  of  these  two 
factors  of  the  multiplier  (9),  obtaining  8874;    and  multiply 


8874          tnig  resuit  by  the  remaining  factor  of  the  multiplier   (7), 
7          obtaining  62118  as  the  required  product.     If  63  =  7  times  9, 


62118          tnen  ^  times  986  must  equal  7  times  9  times  986. 

EXAMPLES  FOR  PRACTISE 
Employing  continued  multiplication,  multiply 

1.  6783  by  72  5.  758439  by  96 

2.  52896  by  49  6.  4273869  by  132 

3.  78348  by  54  7.  3842768  by  144 

4.  925687  by  42  8.  52783  by  432  (=8X9X6) 

30.  If  a  number  be  multiplied  by  1,  the  resulting  product  will 
be  found  equal  to  the  multiplicand.     Hence,  if  the  same  number 
be  multiplied  by  ten  times  1,  or  10,  the  resulting  product  must  be 
ten  times  the  multiplicand,  or  that  multiplicand  with  one  cipher  at 
the  right  of  it  [9,  b] ;  or  if  that  number  be  multiplied  by  one  hun- 
dred times  1,  or  100,  the  resulting  product  must  be  one  hundred 
times  the  multiplicand,  or  that  multiplicand  with  two  ciphers  at 
the  right  of  it  [9,  c],  etc.     Hence,  to  multiply  any  number  by  1, 
followed  by  one  or  more  ciphers,  simply  place  as  many  ciphers  at 
the  right  of  the  multiplicand  as  are  found  at  the  right  of  1  in  the 
multiplier. 

EXAMPLES  FOR  PRACTISE 
Multiply  Multiply 

1.  8763  by  100  4.   62875  by  1000 

2.  9578  by  10000  6.   41329  by  100000 

3.  75623  by  10  6.   72865  by  10000 

31.  If  a  multiplier  be  greater  than  1,  the  product  must  be  as 
many  times  the  multiplicand  as  that  multiplier  is  times  1.     There- 
fore if  the  multiplier  be  8,  the  product  must  be  eight  times  the 
multiplicand,  and  if  the  multiplier  be  ten  times  8,  or  80,  the  result- 


16  MULTIPLICATION 

ing  product  must  be  ten  times  the  preceding  product  by  8,  or 
eight  times  the  multiplicand  with  one  cipher  at  the  right  of  it 
[9,  6];  and  if  the  multiplier  be  one  hundred  times  8,  or  800,  the 
resulting  product  must  be  one  hundred  times  the  product  by  8, 
or  eight  times  the  multiplicand  with  two  ciphers  at  the  right  of  it 
[9,  c],  etc. 

ILLUSTRATIVE  EXAMPLE 
Multiply  876  by  9000. 

SOLUTION  EXPLANATION.     Multiply  the  multiplicand   (876)   by  9, 

as  if  it  stood  alone,  obtaining  7884.     As  the  true  multiplier 

is  one  thousand  times  9.  the  true  product  must  be  one  thou- 
onnn 

sand  times  the  product  by  9,  or  7884  with  three  ciphers  at 


7884000      the  right  of  it  [9,  d]. 

EXAMPLES  FOR  PRACTISE 
Multiply  Multiply 

1.  894  by  700  4.   6792  by  57000 

2.  7683  by  32000  5.   863  by  437000 

3.  5789  by  64900  6.   9187  by  190000 

32.  If,  as  shown  in  111.  Ex.,  31,  the  product  of  876  multi- 
plied by  9000  is  9  times  876,  or  7884,  with  three  ciphers  at 
the  right  of  it  (7884000),  it  must  follow  that  one  hundred  times 
876,  or  87600,  multiplied  by  9000,  must  be  one  hundred  times  the 
former  product,  or  the  former  product  (7884000)  with  two  addi- 
tional ciphers  at  the  right  of  it,  that  is,  788400000. 

ILLUSTRATIVE  EXAMPLE 

Multiply  27300  by  85000. 

EXPLANATION.     So  arrange  the  given  factors  that  the 
right-hand   significant   figure  of  the  multiplier   (5)   shall 
27300  faU  under  the  right-hand  significant  figure  of  the  multipli- 

85000          cand  (3).      Then  multiply  as  if  there  were  no  ciphers  at 
the  right  of  either  factor,  obtaining  23205.     As,  however, 


the  true  multiplier  is  one  thousand  times  85,  the  product 

2184 of  273  multiplied  by  the  true  multiplier  (85000)  must  be 

2320500000     one  tnousancl  times  23205,  or  23205000;    and  as  the  true 
multiplicand  is  one  hundred  times  273,  or  27300,  the  true 
product  of  27300  multiplied  by  85000  must  be  one  hundred  times  23205000, 
or  2320500000.     Hence, 


DIVISION  17 

RULE.  To  multiply  when  either  or  both  factors  terminate 
hi  ciphers:  So  place  the  multiplier  under  the  multiplicand  that  the 
right-hand  significant  figure  of  the  former  shall  fall  under  the  right- 
hand  significant  figure  of  the  latter,  and  multiply  as  if  there  were  no 
ciphers  at  the  right  of  either.  At  the  right  of  the  resulting  product, 
place  as  many  ciphers  as  are  at  the  right  of  both  factors. 

EXAMPLES  FOR  PRACTISE 

Multiply  Multiply 

1.  468000  by  7200  5.   823000  by  47000 

2.  67200  by  9000  6.    12900  by  680000 

3.  5286000  by  30  7.   3400000  by  90000 

4.  9270000  by  2800  8.   7820000  by  5300 

DIVISION 

33.  Division  is  the  name  of  the  process  which  is  used  when  the 
total  of  all  the  considered  equal  quantities  is  given,  and  it  is  re- 
quired to  find  the  value  of  one  of  those  equal  quantities,  or  the 
number  of  those  equal  quantities. 

NOTE  1.  Division  involves  the  reverse  use  of  the  same  terms  as  have 
already  been  considered  in  multiplication  [Notes  1  and  2,  26].  These  terms 
of  multiplication  are  however  changed  in  name  to  accord  with  the  reverse 
process  employed  in  their  use.  (a)  The  total  of  all  the  equal  quantities,  for- 
merly called  product  [Note  1,  c,  26],  is  now  called  dividend  which  means  a  quan- 
tity to  be  divided.  Hence,  one  of  the  equal  quantities  and  the  number  of  such 
equal  quantities  must  be  the  factors  or  producers  of  the  dividend  [Note  2, 
25].  (6)  Either  of  these  two  factors,  if  given,  and  used  to  divide  the  dividend, 
is  distinguishably  called  the  divisor  which  means  "that  by  which  a  quantity 
is  divided";  and  (c)  the  remaining  factor,  resulting  from  the  division,  is 
distinguishably  called  the  quotient,  which  means  "the  number  of  times  one 
quantity  is  contained  in  another."  Hence, 

NOTE  2.  (a)  The  dividend  is  the  number  to  be  divided;  (6)  the  divisor 
is  the  number  by  which  the  dividend  is  divided;  and  (c)  the  quotient  is  the 
result  obtained  from  the  division. 

NOTE  3.  After  the  last  quotient  figure  has  been  obtained,  the  undivided 
portion  of  the  dividend,  if  any,  is  called  the  remainder. 

NOTE  4.  (a)  Division  is  indicated  by  the  sign  -f-  which  follows  the 
dividend  and  precedes  the  divisor.  (6)  It  is  read  divided  by. 

34.  Principles     of    division.     1.   //   the   divisor   and   dividend 
are  concrete  numbers,  they  must  both  represent  units  of  the  same 


18  DIVISION 

name  to  produce  the  true  quotient.  2.  //  the  dividend  be  concrete 
and  the  divisor  abstract,  the  quotient  will  be  concrete  and  express 
units  of  the  same  name  as  the  dividend.  3.  The  remainder,  being 
the  undivided  portion  of  the  dividend  [Note.  3,  33],  must  express 
units  of  the  same  name  as  the  dividend.  4.  To  multiply  or  divide 
the  dividend  by  a  number  is  equivalent  to  multiplying  or  dividing  the 
quotient  by  the  same  number.  5.  To  multiply  or  divide  both  divisor 
and  dividend  by  the  same  number  produces  no  change  in  the  quotient. 
6.  To  multiply  the  divisor  by  a  number,  the  dividend  remaining  the 
same,  is  equivalent  to  dividing  the  quotient  by  the  same  number.  7. 
To  divide  the  divisor  by  a  number,  the  dividend  remaining  the  same, 
is  equivalent  to  multiplying  the  quotient  by  the  same  number.  8. 
The  complete  quotient  equals  the  quotient  of  each  order  of  the  dividend 
separately  divided  by  the  divisor. 

SHORT   DIVISION 

35.  Divide  895  by  5. 

EXPLANATION.     Divide  the  highest  order  of  the  dividend 
SOLUTION    (8  hundred)  by  the  divisor  (5)  to  find  the  highest  order  of  the 
5)895        quotient,  obtaining  1  hundred,  and  leaving  an  undivided  re- 
mainder  of  3  hundred  in  the  dividend.     Write  the  first  quotient 
figure  (1  hundred)  underneath  the  8  hundred  of  the  dividend 
from  which  it  was  obtained. 

Carry  the  3  undivided  hundred  (=30  tens)  of  the  preceding  operation 
to  the  next  lower  order  of  the  dividend  (9  tens),  and  divide  the  result  (39  tens) 
by  the  divisor  (5),  obtaining  7  tens  as  the  next  lower  order  of  the  quotient, 
leaving  4  tens  of  the  39  tens  as  an  undivided  remainder.  Write  this  second 
quotient  figure  (7  tens)  underneath  the  order  of  the  dividend  (9  tens)  from 
which  it  was  obtained. 

Carry  the  4  undivided  tens  (=  40  units)  to  the  next  lower  order  of  the 
dividend  (5  units),  and  divide  the  result  (45  units)  by  the  divisor  (5),  obtain- 
ing exactly  9  units,  as  the  next  lower  order  of  the  quotient.  Write  this  third 
and  final  quotient  figure  (9  units)  underneath  the  units'  order  of  the 
dividend  from  which  it  was  obtained. 

NOTE  1.  The  above  method  of  obtaining  the  successive  quotient 
figures  by  mental  processes  is  called  short  division.  Short  division  is  em- 
ployed when  the  divisor  is  not  greater  than  12. 

NOTE  2.  To  prove  division,  multiply  the  quotient  by  the  divisor,  and  to 
the  product  add  the  remainder,  if  any.  The  result  should  equal  the  dividend 
[Note  1,  33]. 


DIVISION 
EXAMPLES  FOR  PRACTISE 


19 


Divide 

1.  71862496  by  4 

2.  827359281  by  3 

3.  952873656  by  6 

4.  829673845  by  7 

5.  528379186  by  2 

6.  734829635  by  5 


Divide 

7.  48329157684  by  12 

8.  25671928347  by  9 

9.  76982763451  by  11 

10.  82534826780  by  10 

11.  50380700608  by  8 

12.  90070008301  by  7 


LONG  DIVISION 


36.  Divide  74865  by  58. 


SOLUTION 

58)74865(1290j| 
58 
168 
116 


526 
522 
45 


EXPLANATION.  Divide  as  many  of  the  highest 
orders  of  the  dividend  as  will  contain  the  divisor 
(74  thousand)  by  the  divisor  (58),  obtaining  1  thou- 
sand as  the  highest  order  of  the  quotient.  Write 
1  (thousand)  as  the  first  quotient  figure,  multiply 
the  divisor  (58)  by  this  quotient  figure  (1  thousand) 
and  subtract  the  product  (58  thousand)  from  74 
thousand,  thus  finding  that  16  thousand  of  the  74 
thousand  remain  undivided. 

Carry  the  16  undivided  thousands  ( =  160  hun- 
dred) of  the  dividend  to  the  next  lower  order  of  the 


dividend  (8  hundred),  which  is  most  conveniently  done  by  "bringing  down" 
the  8  hundred  of  the  dividend  to  the  right  of  the  16  undivided  thousand,  ob- 
taining 168  hundred  as  the  next  portion  of  the  dividend  to  be  divided,  and 
168  hundred  -5-  58  equal  2  hundred,  the  next  quotient  figure.  Multiply  the 
divisor  (58)  by  this  second  quotient  figure  (2  hundred),  and  subtract  the 
product  (116  hundred)  from  168  hundred,  thus  finding  that  52  hundred  of 
the  168  hundred  remain  undivided. 

Carry  the  52  undivided  hundreds  ( =  520  tens)  of  the  last  considered  part 
of  the  dividend  to  the  next  lower  order  of  the  dividend  (6  tens),  by  "bring- 
ing down"  the  6  tens  of  the  dividend  to  the  right  of  the  52  hundreds,  obtaining 
526  tens  as  the  next  portion  of  the  dividend  to  be  divided;  and  526  tens  -5- 
58  =  9  tens.  Write  9  tens  as  the  next  quotient  figure,  multiply  the  divisor 
(58)  by  this  third  quotient  figure  (9  tens)  and  subtract  the  product  (522  tens) 
from  526  tens,  thus  finding  that  4  tens  of  the  526  tens  remain  undivided. 

Carry  the  4  undivided  tens  (=  40  units)  of  the  considered  part  of 
the  dividend  to  the  next  lower  order  of  the  dividend  (5  units),  and  divide 
the  result  (45  units)  by  58,  obtaining  0  units  times,  thus  finding  that  58 
are  contained  in  74865,  1290  times,  and  that  45  is  the  final  undivided 
portion  of  the  dividend.  Write  the  divisor  58  under  this  undivided  part 
of  the  dividend  to  denote  this  unexecuted  part  of  the  division,  and  place 


20  DIVISION 

this  fractional  part  of  the  quotient  (ff)  to  the  right  of  the  integral  part 
(1290),  obtaining  1290ff  as  the  complete  quotient. 

NOTE  1.  The  above  method  of  obtaining  the  successive  quotient  figures 
by  written  processes  is  called  long  division.  Long  division  is  usually  em- 
ployed when  the  divisor  is  greater  than  12. 

NOTE  2.  (a)  To  find  the  first  quotient  figure  when  the  divisor  consists 
of  several  figures,  mentally  divide  the  first  figure  or  the  first  two  figures  at 
the  left  of  the  dividend  by  the  first  figure  at  the  left  of  the  divisor,  or  by  the 
first  figure  plus  1  if  the  second  figure  from  the  left  of  the  divisor  is  greater 
than  5.  (fc)  To  find  each  succeeding  quotient  figure,  point  off  from  the  right 
of  each  partial  dividend  as  many  figures  lacking  one  as  there  are  figures  in 
the  divisor,  and  divide  the  remaining  figures  of  the  partial  dividend  by  the 
left-hand  figure  of  the  divisor  as  in  (a). 

NOTE  3.  If,  after  multiplying  the  divisor  by  the  last  obtained  quotient 
figure,  the  product  is  found  to  be  greater  than  the  partial  dividend  above  it, 
the  last  quotient  figure  is  too  great  and  should  be  diminished  by  1. 

NOTE  4.  If,  after  any  subtraction  from  a  partial  dividend,  a  remainder 
is  obtained  which  is  greater  than  the  divisor,  the  last  obtained  quotient  figure 
is  incorrect,  and  should  be  increased  by  1. 


EXAMPLES  FOR  PRACTISE 

Divide  Divide 

1.  876349572  by  23  10.   4378692573  by  486 

2.  924183762  by  41  11.   5827394618  by  928 

3.  729385647  by  63  12.   6537428719  by  4365 

4.  682753725  by  37  13.   8291736451  by  72849 
6.   521837649  by  58  14.   5278060072  by  603005 

6.  437529183  by  79  15.   9428910003  by  412078 

7.  621938726  by  48  16.   8006000342  by  890065 

8.  529486312  by  67  17.   7200650004  by  27384 

9.  297834263  by  325  18.   4120082005  by  728163 

37.  A  continued  division  is  a  successive  division  by  two  or 
more  divisors.  It  is  usually  employed  when  an  inconvenient 
divisor,  as  48,  involving  long  division,  can  be  changed  to  two  or 
more  convenient  divisors  (48  =  8  X  6),  involving  successive 
short  divisions. 

ILLUSTRATIVE  EXAMPLE 
Divide  6487  by  336. 


DIVISION 


21 


SOLUTION 

336  =  8  X  7  X  6 

8)6487 
7)810  +  7  = 
6)115  +  5  (X  8)  = 
19  +  1 


Quotient 


EXPLANATION.     If  6487  be  divided  by 
8  or  one  forty-second  of  the  true  divisor, 
the  quotient   (810)   will  be  42  times  the 
true  quotient  [Prin.  7,  34],  and  a  remainder 
of  7.     As  this  remainder  is  from  a  partial 
division  of  the  true  dividend  it  must  form 
•7    a  part  of  the  true  remainder  [Prin.  3,  34]. 
If  810  (forty-two  times  the  true  quo- 
~    tient)  be  divided  by  42,  the  result  will  be 
8  X  7)  =56    the  true  quotient;   but  if  810  be  divided 
by  the  second  factor  7,  or  one-sixth  of  42, 
the   resulting   quotient  (115)    must   be   6 
times  the  true  quotient  [Prin.  7,  34]  and 
a  remainder  of  5.     As  this  remainder  (5) 

is  from  a  partial  division  of  one-eighth  of  the  true  dividend  (810  =  one- 
eighth  of  6487),  it  must  be  one-eighth  of  the  second  part  of  the  true  re- 
mainder [Prin.  3,  34]  and  8  times  5  or  40  will  form  the  second  part  of  the 
true  remainder. 

As  115  is  6  times  the  true  quotient,  divide  115  by  6,  the  third  factor,  to 
find  the  true  quotient  (19),  and  a  remainder  of  1.  As  this  remainder  (1)  is 
from  a  division  of  one-seventh  of  one-eighth  of  the  true  dividend,  it  must 
be  one-seventh  of  one-eighth  of  the  true  remainder  [Prin.  3,  34],  and  7  times 
8  times  1,  or  56,  will  form  another  part  of  the  true  remainder.  Hence,  add 
these  several  parts  of  the  true  remainder,  obtaining  7  +  40  +  56,  or  103  as 
the  complete  remainder. 

SUMMARY.  1.  The  successive  division  by  the  factors  of  a  divisor 
is  equivalent  to  a  division  by  that  divisor.  2.  The  complete  remain- 
der of  a  successive  division  by  the  factors  of  a  divisor  equals  the  first 
remainder  -\-the  product  of  the  second  remainder  multiplied  by  the 
first  divisor  +  the  continued  product  of  any  succeeding  remainder 
multiplied  successively  by  all  the  divisors  which  precede  its  own 
divisor. 


PRACTISE 

Divide 

3678251968  by  96 
4826731853  by  84 
2372853  by  168  (8X7X3) 
7283945  by  504  (9X8X7) 
5678325  by  125  (5X5X5) 
9671847  by  315  (9X7X5) 


EXAMPLES 

FOR 

Divide 

1. 

896542  by  28 

7. 

2. 

1768329  by  32 

8. 

3. 

768296854  by  45 

9. 

4. 

93725823  by  72 

10. 

5. 

62834251  by  64 

11. 

6. 

5162738429  by  54 

12. 

22  DIVISION 

38.  Since  the  value  of  an  order  [9,  g],  is  decreased  tenfold  for 
each  place  that  it  is  moved  to  the  right  [9,  /],  if  all  the  orders  of  a 
dividend  be  moved  one  place  to  the  right,  it  will  be  equivalent  to 
dividing  that  dividend  by  10;    if  moved  two  places  to  the  right 
it  will  be  equivalent  to  dividing  that  dividend  by  100;   if  three 
places,  by  1000,  etc.     Now  to  strike  off  one  figure  from  the  right 
of  any  dividend  will  have  the  effect  of  moving  its  remaining  orders 
one  place  further  to  the  right  than  their  original  position,  and 
thus  to  dividing  that  dividend  by  10;  to  strike  off  two  figures  will 
move  the  remaining  orders  two  places  to  the  right  of  their  original 
position,  and  thus  to  dividing  by  100,  etc.     It  also  follows  that  the 
figure  or  figures  which  are  taken  from  the  right  of  the  dividend 
will  still  retain  the  same  local  position  and  value  which  they  had 
in  the  original  dividend,  and  must  be  the  undivided  portion  of 
the  original  dividend,  or  the  true  remainder.     Hence, 

RULE.  To  divide  by  10,  100,  1000,  etc.,  cut  off  from  the  right  of 
the  dividend  as  many  figures  as  there  are  O's  at  the  right  of  1  in  the 
divisor.  The  remaining  figures  of  the  dividend  will  constitute  the 
quotient;  and  the  figures  cut  off  will  be  the  remainder. 

EXAMPLES  FOR  PRACTISE 
Divide  Divide 

1.  67859  by  10  6.   1483762  by  1000 

2.  82678  by  100  6.   7834106  by  100 

3.  92567  by  1000  7.   28710015  by  10000 

4.  26835  by  10000  8.   62520003  by  1000 

39.  If,  as  found  in  38,  when  three  figures  are  stricken  from  the 
right  of  a  number,  the  remaining  figures  will  be  the  true  quotient 
of  that  number  divided  by  1000,  and  the  figures  stricken  off  will 
be  the  true  remainder,  it  must  follow  that  striking  off  three  figures 
from  the  right  of  that  number  will  be  seven  times  the  true  quotient 
of  that  number  divided  by  7  times  1000,  or  7000  [Prin.  7,  34]  and, 
therefore,  that  striking  off  three  figures  from  a  number  and  divid- 
ing its  remaining  figures  by  7,  must  be  the  quotient  of  that  number 
divided  by  7000.     Further,  if,  as  also  shown  in  38,  the  three  figures 
stricken  from  the  right  of  the  dividend  constitute  the  true  remain- 
der from  dividing  by  1000,  they  must  also  form  a  part  of  the  true 


DIVISION  23 

remainder  from  dividing  by  seven  times  1000  [Prin.  3,  34];  and 
any  succeeding  remainder  from  dividing  the  remaining  figures  of 
the  dividend  (or  one-thousandth  of  the  true  dividend)  by  7  will 
be  one-thousandth  of  the  second  part  of  the  true  remainder,  and 
hence  must  have  three  figures  at  its  right  to  exhibit  its  true  local 
value  [9,  d].  If,  therefore,  the  three  figures  cut  off  from  the  right 
of  the  original  dividend,  which  constitute  one  part  of  the  true  re- 
mainder, be  placed  at  the  right  of  the  second  remainder,  the  result 
will  be  the  complete  true  remainder. 

RULE.  When  any  divisor  has  ciphers  at  its  right:  1.  Cut  off 
from  the  right  of  the  dividend  as  many  figures  as  there  are  O's  at  the 
right  of  the  divisor.  2.  Divide  the  remaining  figures  of  the  dividend 
by  the  divisor  without  the  O's  at  its  right,  to  find  the  true  quotient. 
3.  Place  the  figures  cut  off  from  the  original  dividend  at  the  right  of 
the  final  remainder  to  obtain  the  complete  remainder. 

ILLUSTRATIVE  EXAMPLE 

Divide  93865  by  3700. 

EXPLANATION.    3700  =  100  X  37.    To 
divide  93865  by  the  first  factor  (100),  cut 

37p£))938£>;)(25  Quo.  off  two  fig1""68  from  tne  rignt  of  tne  divi- 

dend, obtaining  938  as  37  times  the  true 
quotient,  and  65  as  part  of  the  true  remain- 
der [38].  Then, 

185  Divide  938  by  37,  obtaining  the  true 

~~Ts65  "Rpm  quotient  (25)  and  a  second  remainder  of 

13.  As  this  remainder  (13)  is  from  divid- 
ing one-hundredth  of  the  original  dividend  (93865),  it  is  one-hundredth  of 
the  true  remainder  [Prin.  3,  34],  and  100  times  13,  or  13  hundred  must  be 
the  second  part  of  the  true  remainder.  Hence,  place  the  first  part  of  the 
true  remainder  (65)  to  the  right  of  the  second  part  (13  hundred),  obtaining 
1365  as  the  complete  remainder. 

EXAMPLES  FOR  PRACTISE 
Divide  Divide 

1.  8736541  by  800    6.  73289653  by  218000 

2.  9183762  by  6000    7.  52671839  by  493000 

3.  4275918  by  400    8.  42835267  by  63000  (1000  X  9  X  7) 

4.  6823475  by  7000    9.  84276834  by  4800  (100  X  8  X  6) 

6.  5938764  by  4100   10.  64837291  by  350000  (10000  X  7  X  5) 


24  UNITED   STATES    MONEY 

UNITED  STATES  MONEY 

40.  United  States  money  is  usually  expressed  in  dollars  and 
cents;   but  in  performing  calculations  in  United  States  money,  it 
has  been  found  convenient  in  certain  cases  to  be  more  specific  in 
designating  its  orders  by  naming  the  first  figure  to  the  right  of  the 
decimal  point,  dimes,  and  the  third  figure  to  the  right  of  the  deci- 
mal point,  mills. 

NOTE  1.  Scale:  10  mills  =  1  cent;  10  cents  =  1  dime;  10  dimes  = 
1  dollar. 

NOTE  2.  In  expressing  the  several  orders  of  United  States  money,  the 
dollars  are  written  in  accordance  with  9.  A  point  (.)  is  used  to  separate  the 
dollars  from  the  lower  orders.  The  first  order  to  the  right  of  the  point 
expresses  dimes;  the  first  two  orders  to  the  right  of  the  point,  considered  as  one 
number,  express  cents;  and  the  third  order  to  the  right  of  the  point  expresses 
mills. 

NOTE  3.  To  indicate  that  a  certain  number  expresses  dollars,  the  sign  $ 
is  placed  at  its  left;  and  that  a  certain  number  expresses  cents,  the  sign  £  is 
placed  at  its  right. 

NOTE  4.  If  the  cents  to  be  expressed  are  less  than  10,  a  cipher  should 
be  written  after  the  point  to  denote  the  absence  of  tens  of  cents  or  dimes. 
Thus,  8  dollars  7  cents  should  be  written  $8.07. 

EXERCISES 
Write  Write 

1.  78  dollars,  19  cents  8.  16  dollars,  8  dimes 

2.  267  dollars,  23  cents  9.  341  dollars,  6  mills 

3.  59  dollars,  42  cents  10.  472  dollars,  2  cents 

4.  167  dollars,  3  cents  11.  673  dollars,  34  cents,  7  mills 

5.  5273  dollars,  8  cents  12.  9183  dollars,  7  cents,  2  mills 

6.  435  dollars,  36  cents  13.  4865  dollars,  5  mills 

7.  2926  dollars,  5  cents  14.  793  dollars,  1  cent,  9  mills 

Express  the  following  by  spoken  or  written  words : 

15.  $35.42          18.     $500.34          21.   $86.352          24.     $80.706 

16.  $76.28          19.   $2176.09          22.   $75.089          25.   $625.105 

17.  $95.05          20.     $513.06          23.   $54.003          26.   $532.017 

41.  (a)  As  10  dimes,  or  100  cents,  or  1000  mills  equal  1  dollar 
[Note  1,  40],  dollars  may  be  changed  to  dimes  by  annexing  one 


UNITED   STATES    MONEY  25 

cipher,  to  cents  by  annexing  two  ciphers,  and  to  mills  by  annex- 
ing three  ciphers  [30].  (b)  As  10  cents  or  100  mills  equal  1  dime, 
dimes  may  be  changed  to  cents  by  annexing  one  cipher,  and  to 
mills  by  annexing  two  ciphers,  (c)  Similarly  cents  may  be  changed 
to  mills  by  annexing  one  cipher,  (d)  The  lower  orders  may  be 
changed  to  dollars  by  pointing  off  one  figure  from  the  right  of 
the  dimes,  two  figures  from  the  right  of  the  cents,  or  three  figures 
from  the  right  of  the  mills  [38],  prefixing  ciphers  if  necessary. 

EXERCISES 

Change  Change 

1.  $748  to  cents  6.  8  cents  to  mills 

2.  $57  to  mills  7.  5278  cents  to  dollars 

3.  $382  to  cents  8.  6175  mills  to  cents 

4.  7  dimes  to  cents  9.  69183  cents  to  dollars 
6.  3  dimes  to  mills  10.  5368  dimes  to  dollars 

ADDITION  OF  UNITED  STATES  MONEY 

42.  The   processes   for   adding,   subtracting,   or   multiplying 
United  States  money  are  the  same  as  for  abstract  numbers,  both 
being  expressed  upon  a  scale  in  which  10  units  of  any  order  equal 
1  unit  of  the  next  higher  order  [compare  Note  1,  40  with  e  and/, 
9],     Any  expression  in  United  States  money  may  therefore  be 
considered  a  simple  integral  number  of  the  same  name  as  its  right- 
hand  order,  and  added  as  such,  and  the  resulting  sum  changed 
back  to  dollars  by  d,  41 . 

ILLUSTRATIVE  EXAMPLE 

43.  Add  $372,  $631.18,  $9352.605,  and  $43. 

EXPLANATION.     So  arrange  the  numbers  to  be  added 

SOLUTION  that  the  p0ints  of  each  shall  fall  in  an  erect  column,  thus 

<£     372  causing  mills  to  fall  under  mills,  cents  under  cents,  dimes 

„*•,  -,  o  under  dimes,  dollars  under  dollars  [18].     Add  and  carry 

as  in  ordinary  addition  [19].     As  the  sum  of  the  dimes 

9352.605  column  (6  +  1)  must  be  7  dimes,  and  the  sum  of  the 

43.  dollars'  column  (3+2  +  1+2)  must  be  8  dollars,  a 

$10398  785  point  should  be  placed  in  the  sum  between  the  7  dimes 

and  8  dollars  to  denote  their  respective  values  [Note  2, 

40],  that  is,  the  point  in  the  sum  should  fall  directly  under  the  column  of  points 

in  the  numbers  added. 


26  UNITED    STATES    MONEY 

EXAMPLES  FOR  PRACTISE 

1.  Add  $573.25,  $6187.92,  $48.34,  $7283.16,  $8.64,  and  $58.25. 

2.  Add  $589.18,  $39.42,  $643.28,  $7.24,  $826.32,  and  $65.58. 

3.  Add  $82.74,  $9.36,  $628.15,  $97.87,  $527.62,  and  $4.50. 

4.  Add  3  dollars,  4  cents;   28  dollars,  thirty  cents;   7  dollars, 
2  mills;  45  dollars,  9  mills;  19  dollars,  5  cents;  356  dollars,  8  cents. 

5.  Add  9  cents;  2  dollars,  18  cents;  74  cents;  3  cents,  5  mills; 
4  dollars,  25  cents;  2  cents;  34  cents;  4  dollars;  15  dollars,  5  cents. 

SUBTRACTION  OF  UNITED  STATES  MONEY 
ILLUSTRATIVE  EXAMPLE 

44.  Subtract  $8,  7^  from  $22,  15^ 

SOLUTION  EXPLANATION.  So  place  the  subtrahend  (8.07)  that  the 

point  at  the  right  of  its  dollars  shall  fall  under  the  point  at 

$22.15  the  right  of  the  dollars  in  the  minuend  (22.15),  thus  causing 

g  07  similar  orders  to  fall  in  the  same  column.     Subtract,  borrow, 

and  carry  as  in  ordinary  subtraction  [23],  and  place  a  point 

in  the  remainder  directly  under  those  of  the  minuend  and 

subtrahend,  as  already  explained  in  pointing  off  the  sum  in  111.  Ex.,  43. 

EXAMPLES  FOR  PRACTISE 

Subtract  Subtract 

1.  $518.25  from  $838.13       5.   $358,  5ff  from  $500 

2.  $28.32  from  $47.157         6.   $67  from  $285,  80^ 

3.  $813  from  $1826.45          7.   $5,  4^f  from  $12,  3  mills 

4.  $534.68  from  $913  8.   $248  from  $326,  5^ 

MULTIPLICATION   OF  UNITED  STATES   MONEY 
ILLUSTRATIVE  EXAMPLE 

45.  Multiply  $873.42  by  53. 

SOLUTION  EXPLANATION.     Write    the    multiplier    under    the    mul- 

tiplicand  so  that   the  right-hand   significant  figure  of  the 
$873.42        multiplier  shall  fall  under  the  right-hand  significant  figure 
53        of  the  multiplicand.     Multiply  and  carry  as  in  ordinary  mul- 
262026        tiplication  [28].     As  the  multiplicand  is  equivalent  to  87342 
4^710          cents,  53  times  87342  cents  must  also  be  cents  [Prin.  1,  26]. 
Hence,  change  the  resulting  product  4629126  cents  to  dollars 


$46291.26        as  shown  in  d,  41,  that  is,  point  off  from  the  right  of  the  product 
as  many  figures  as  are  pointed  off  from  the  right  of  the  multiplicand. 


DIVISION    OF   UNITED    STATES    MONEY  27 

EXAMPLES  FOR  PRACTISE 

Multiply  Multiply 

1.  $6875  by  253  6.  $357  and  16^  by  72  (8  X  9) 

2.  $768.25  by  48  (8  X  6)  7.  $213  and  Si  by  258 

3.  $27.53  by  37  8.  $562,  3?f,  4  mills  by  24 

4.  $827.50  by  468  9.  $65,  9  mills  by  900 

6.   $91  and  3fi  by  53  10.  $8,  15^,  7  mills  by  32000 

DIVISION  OF  UNITED  STATES  MONEY  BY  AN  ABSTRACT 

DIVISOR 

46.  In  division  of  United  States  money,  the  dividend  may  be 
regarded  as  a  simple  integer  of  the  same  name  as  its  right-hand  or 
lowest  order.  Thus,  $5.26  may  be  regarded  as  526  cents,  and 
$2.065  as  2065  mills.  Hence,  if  the  divisor  is  an  abstract  number, 
the  process  of  division  can  differ  in  no  respect  from  ordinary 
division  as  shown  in  36,  and  the  resulting  quotient  must  express 
units  of  the  same  name  as  the  dividend  [Prin.  2,  34]. 

ILLUSTRATIVE  EXAMPLE 
Divide  $593.95  by  47. 

SOLUTION  EXPLANATION.     Divide  as  in  ordinary  division. 

4  As  the  dividend  expresses  59395  cents,  one  forty- 

A7^p;QQ  Qf^«l9  A3  seventh  of  the  dividend,  or  1264  must  also  express 

^™*'  cents  [Prin.  2,  34].     Hence  change  the  cents  (1264) 

47  to  dollars  by  cutting  off  two  figures  from  the  right 

123  of  the  quotient  [41,  d]  obtaining  $12.64. 

Q4  As  there  is  no  coinage  in  U.  S.  money  below  the 

cents'  order,  it  is  customary  to  stop  the  division 

after  bringing  down  and  dividing  the  cents'  order 

282  of  the  dividend,  and  if  the  final  remainder  be  one- 

175  half  the  divisor,  or  more,  to  increase  the  last  figure 

..  ,-  of  the  quotient  by  1.     Hence,  as  34,  the  final  re- 

mainder  is  more  than  half  the  divisor  (47),  the  last 

34  figure  of  the  quotient  (3)  is  increased  by  1. 

NOTE.     If  the  dividend  is  expressed  in  dollars 

only,  and  it  should  not  be  exactly  divisible  by  the  divisor,  change  the  divi- 
dend to  cents  [41,  a]  by  annexing  two  ciphers,  and  continue  the  division 
until  these  annexed  ciphers  have  been  brought  down  and  divided.  The 
resulting  quotient  will  then  express  cents. 


28  DIVISION    OF   UNITED    STATES    MONEY 

CAUTION.  If  the  division  is  not  the  final  operation  of  a  solution,  but  is 
to  be  followed  by  an  operation  in  multiplication,  it  will  be  necessary  to  retain 
the  exact  remainder  and  express  it  as  a  fraction  of  a  cent  as  in  111.  Ex.  36; 
for  any  omitted  inconsiderable  fraction  of  a  cent  if  multiplied  by  a  con- 
siderable multiplier  may  cause  an  error  in  the  final  answer  of  many  integral 
cents. 

EXAMPLES  FOR  PRACTISE 

Divide  Divide 

1.  $6783.25  by  5  5.  $763  by  19 

2.  $786.97  by  8  6.  $578  by  83 

3.  $428.34  by  23  7.  $2783.42  by  327 

4.  $562.78  by  53  8.  $8675  by  465 

DIVISION  OF  UNITED  STATES  MONEY  BY  UNITED  STATES 

MONEY 

47.  When  both  the  divisor  and  dividend  express  United  States 
money,  and  they  are  regarded  as  integers  of  the  same  name  as 
their  respective  right-hand  orders,  they  must  first  be  changed  to 
express  units  of  the  same  name  if  their  right-hand  orders  are  dis- 
similar [Prin.  1,  34].  This  is  accomplished  by  annexing  the  neces- 
sary number  of  ciphers  to  that  term  which  has  the  fewer  figures 
at  the  right  of  the  point,  so  that  the  right-hand  orders  of  both  the 
divisor  and  the  dividend  shall  be  equidistant  from  their  respective 
points. 

ILLUSTRATIVE  EXAMPLE 
Divide  $6273  by  $18.25. 

SOLUTION 

$18.25)$6273.00(343  f|M  EXPLANATION.  As  the  divisor 
_  7_  expresses  1825  cents,  reduce  the  divi- 
dend  ($6273)  to  cents  by  annexing 

7980  two   ciphers  [Prin.   1,  34].      Divide 

7300 

as  in  ordinary  division,  finding  that 


6800  1825  cents  are  contained  in  627300 

5475  Cents  343  iff!  times. 

1325 


RELATION    OF    NUMBERS  29 

EXAMPLES  FOR  PRACTISE 
Divide  Divide 

1.  $178.32  by  $.03  6.  $9258.63  by  18ff 

2.  $2735.15  by  $2.83  7.  $62583  by  9ff 

3.  $6825.42  by  $17.38  8.  $43897.38  by  $27 

4.  $48578  by  $52.67  9.  $59687.42  by  $826 

5.  $89264  by  $128.30  10.  $18273  by  $145 

RELATION    OF    NUMBERS 

48.  (a)  All  arithmetical  operations  include  only  four  elemen- 
tary mechanical  processes  —  addition,  subtraction,  multiplication, 
and  division;  and  the  intelligent  solution  of  any  problem,  however 
complicated,  depends  upon  a  knowledge  of  the  conditions  to  which 
each  of  these  processes  is  invariably  limited.  (6)  The  only  con- 
ceivable manner  in  which  a  given  numerical  expression  of  a  quan- 
tity can  be  operated  upon  is  by  adding  another  quantity  to  it 
(increasing  it  by  addition  or  by  multiplication),  or  by  taking  a 
quantity  from  it  (decreasing  it  by  subtraction  or  by  division). 
The  arithmetical  law  of  increase  or  decrease  is  uniform  in  appli- 
cation, and  may  be  stated  as  f ollows : 

(c)  A  given  quantity  is  increased  by  addition  if  the  quantity  or 
quantities  by  which  it  is  required  to  be  augmented  are  unequal  to 
the  given  quantity,  or  to  each  other;  or  (d)  by  multiplication  if 
the  quantity  or  quantities  by  which  it  is  required  to  be  augmented 
are  equal  to  the  given  quantity  and  to  each  other,  (e)  The  result 
of  the  increase  is  distinguishably  called  the  sum  if  the  augmenta- 
tion has  been  by  unequal  quantities;  or  (/)  the  product  if  the 
augmentation  has  been  by  equal  quantities,  (g)  Each  unequal 
numerical  quantity  employed  in  obtaining  a  sum  is  distinguishably 
called  a  component  of  that  sum ;  and  (h)  one  of  the  equal  numerical 
quantities  employed  in  obtaining  a  product  is  called  its  multipli- 
cand, or  quantity  to  be  repeated;  and  (i)  the  number  of  such  equal 
quantities  is  called  its  multiplier,  or  times  the  equal  quantity  is  to 
be  repeated. 

(k)  A  given  quantity  is  decreased  by  recognizing  it  as  a  total  of 
two  or  more  quantities;  and  this  total  is  distinguishably  called  a 


30  RELATION    OF    NUMBERS 

minuend  if  it  is  the  total  of  unequal  quantities;  or  a  dividend  if  it 
is  the  total  of  equal  quantities. 

(m)  Though  all  increases  can  be  effected  by  addition,  the 
process  of  addition  is  usually  limited  to  such  increases  as  are 
maole  by  combining  unequal  quantities  [17],  and  the  process  of 
multiplication  is  more  conveniently  employed  in  combining  equal 
quantities  [25].  (n)  While  all  decreases  can  be  accomplished  by 
subtraction,  the  process  of  subtraction  is  usually  limited  to  the 
decrease  of  such  quantities  as  express  a  total  of  unequal  com- 
ponents [21],  and  the  process  of  division  is  more  conveniently 
employed  in  the  decrease  of  quantities  which  express  a  total  of 
equal  components  [33].  Hence, 

PKINCIPLES.  1.  To  find  the  total  of  two  or  more  given  unequal 
quantities,  employ  addition.  2.  To  find  the  total  of  two  or  more 
given  equal  quantities,  employ  multiplication.  3.  To  find  one 
quantity  of  a  given  total  of  two  or  more  unequal  quantities,  employ 
subtraction.  4.  To  find  one  quantity  of  a  given  total  of  two  or  more 
equal  quantities,  employ  division. 

NOTE  1.  It  is  thus  seen  that  addition  and  subtraction  are  opposite 
processes  pertaining  to  unequal  quantities,  and  that  multiplication  and  divi- 
sion are  opposite  processes  pertaining  to  equal  quantities;  also,  that  addition 
and  multiplication  are  different  processes  for  obtaining  a  required  total,  the 
former  limited  to  unequal  and  the  latter  to  equal  quantities;  and  that  sub- 
traction and  division  are  different  processes  for  diminishing  a  given  total,  the 
former  by  one  or  more  unequal,  and  the  latter  by  one  or  more  equal  quantities. 

NOTE  2.  It  is  also  seen  that  the  successive  steps  in  the  solution  of  any 
problem  are:  1.  Identifying  its  given  numerical  terms.  2.  Ascertaining  the 
relation  of  these  terms  to  each  other.  3.  Perfecting  that  relation  by  per- 
forming the  appropriate  operation  [Note,  7]. 

49.  (a)  The  total  of  two  or  more  unequal  quantities  is  dis- 
tinguishably  called  the  sum  if  required,  or  (b)  the  minuend  if 
given,  (c)  The  total  of  two  or  more  equal  quantities  is  distin- 
guishably  called  the  product  if  required,  or  (d)  the  dividend,  if 
given,  (e)  The  several  unequal  quantities  which  constitute  a 
total  are  called  its  components  if  the  total  (sum)  is  required,  or  (/) 
they  are  called  its  subtrahend  and  its  remainder  if  the  total  (minuend) 
is  given,  (g)  If  a  total  of  two  or  more  equal  quantities  (product) 
is  required,  one  of  the  several  equal  quantities  which  constitute 


RELATION    OF    NUMBERS  31 

the  required  total  is  distinguishably  called  its  multiplicand  factor, 
and  (h)  the  number  of  times  the  equal  quantity  is  to  be  repeated 
is  called  its  multiplier  factor,  (i)  If  a  total  of  two  or  more  equal 
quantities  (dividend)  is  given,  that  one  of  its  two  factors  just 
explained  in  g  and  h,  which  is  given  with  it,  is  distinguishably 
called  its  divisor;  and  (j)  its  other  factor  [Note  1,  33]  which  will 
then  be  required,  is  called  its  quotient. 

50.   Addition  and  subtraction  contrasted. 

ILLUSTRATIVE  EXAMPLES 

1.  J.  A.  Smith  and  S.  H.  Brown  have  formed  a  copartner- 
ship, Smith  investing  $5000  and  Brown  $7000.     What  is  the  cap- 
ital of  the  firm? 

2.  The  capital  of  the  firm  of  Smith  and  Brown  is  $12000. 
If  Smith's  investment  is  $5000,  how  much  is  Brown's? 

SOLUTIONS 

(  $5000  )   Components  of  a 
Components  of  a  given  minuend     •{  . 

(    7000  )       required  sum 

Minuend,  if  given  =  $12000  =  Sum,  if  required. 

EXPLANATION.  The  total  of  two  or  more  unequal  quantities  is  the  sum 
if  required  as  in  Ex.  1,  or  the  minuend  if  given  as  in  Ex.  2  [a  and  6,  49],  Hence, 

In  Ex.  1,  add  the  given  unequal  quantities,  obtaining  $12000  as  the 
required  sum  or  capital  [Prin.  1,  48]. 

In  Ex.  2,  subtract  the  given  unequal  quantity  ($5000),  now  called  the 
subtrahend,  from  the  given  total  of  two  unequal  quantities  ($12000),  now 
called  the  minuend,  to  obtain  the  remaining  unequal  quantity  ($7000),  now 
called  the  remainder  [Prin.  3,  48], 

SUMMARY.  1.  If  a  total  of  two  or  more  unequal  quantities  is 
required,  add  the  several  unequal  quantities.  2.  //  a  total  of  two 
unequal  quantities  is  given,  diminish  it  by  its  given  unequal  quantity 
to  find  its  remaining  unequal  quantity.  3.  If  a  total  of  more  than 
two  unequal  quantities  is  given,  diminish  it  by  the  sum  of  all  its 
unequal  quantities  except  one,  and  the  remainder  will  be  its  one 
remaining  unequal  quantity. 


32  RELATION    OF    NUMBERS 

51.   Multiplication  and  division  contrasted. 
ILLUSTRATIVE  EXAMPLES 

1.  What  is  the  cost  of  12  pounds  of  butter  at  35 £  per 
pound? 

2.  If  12  pounds  of  butter  cost  $4.20,  what  is  the  price  of 
one  pound? 

3.  At  35^f  per  pound,  how  many  pounds  of  butter  can  be 
bought  for  $4.20? 

SOLUTIONS 

Factors  of  a  given  dividend  \      "     ?  Factors  of  a  required  product. 
Dividend,  if  given  =  $4.20  =  Product,  if  required. 

EXPLANATION.  That  factor  of  a  product  which  expresses  one  of  the 
considered  equal  quantities  (35  j£)  is  the  multiplicand  [Note  1,  a,  25] ;  and 
that  factor  which  expresses  the  number  of  such  equal  quantities  (12)  is  the 
multiplier  [Note  1,  b,  25] ;  and  the  total  of  all  the  considered  equal  quantities 
($4.20)  is  the  product  if  required  [Note  1,  c,  25];  or  the  dividend  if  given 
[Note  1,  33].  Hence, 

In  Ex.  1,  multiply  one  of  the  equal  quantities  or  multiplicand  (35  f)  by 
the  number  of  such  quantities  or  multiplier  (12),  obtaining  the  required  total 
of  all  the  equal  quantities  or  product  ($4.20). 

In  Ex.  2,  divide  the  given  total  of  all  the  equal  quantities  or  product 
($4.20),  now  called  the  dividend,  by  the  number  of  such  equal  quantities  or  its 
given  multiplier  factor  (12),  now  called  the  divisor,  obtaining  one  of  the  re- 
quired equal  quantities  or  its  multiplicand  factor  (35^),  now  called  the  quotient 
[Note  1,  33]. 

In  Ex.  3,  divide  the  given  total  of  all  the  equal  quantities  or  product 
($4.20),  now  called  the  dividend,  by  one  of  the  equal  quantities  or  its  given 
multiplicand  factor  (35^),  now  called  the  divisor,  obtaining  the  required 
number  of  such  equal  quantities  or  multiplier  factor  (12),  now  called  the 
quotient. 

SUMMARY.  1.  If  a  product  is  required,  multiply  its  two  given 
factors.  2.  If  a  product  is  given,  divide  it  by  its  given  factor,  and 
the  quotient  will  be  its  other  factor. 


RELATION   OF    NUMBERS  33 

52.   Addition  and  multiplication  contrasted. 
ILLUSTRATIVE  EXAMPLES 

1.  A  farm  consists  of  three  fields,  the  first  containing  275 
acres,  the  second  342  acres,  and  the  third  378  acres.     How  many 
acres  does  the  farm  contain? 

2.  A  farm  consists  of   three   fields,  each   containing  468 
acres.     How  many  acres  does  the  farm  contain? 

SOLUTIONS  EXPLANATION.     In  each  example,  the  same 

(1)  (2)  question  is  asked,  but  the  proper  process  for 

275  4(jg  Ex.  1  is  addition,  because  the  several  components 

n.n  v  ^  °^  ^ne  required  total  are  unequal  [Prin.  1,  48]; 

and  for  Ex.  2  is  multiplication,   because  the 
378  1404  acres     several  components  of  the  required  total  are 

995  acres  ec*ual-    tprin-  2>  48^ 

SUMMARY.  1.  Employ  addition  to  find  the  total  of  two  or  more 
unequal  quantities.  2.  Employ  multiplication  to  find  the  total  of 
two  or  more  equal  quantities. 

NOTE.  The  multiplier  factor,  denoting  the  number  of  equal  quantities 
is  peculiar  to  multiplication  and  division  which  relate  exclusively  to  equal 
quantities.  The  multiplier  factor  cannot  have  its  counterpart  in  addition 
or  subtraction  because  in  each  of  these  processes  the  number  of  unequal  quan- 
tities is  altogether  fortuitous,  depending  upon  the  extent  of  the  inequality  of  the 
several  components  of  a  given  or  required  total. 

53.   Subtraction  and  division  contrasted. 

ILLUSTRATIVE  EXAMPLES 

1.  How  many  acres  will  remain  after  selling  one  field  of 
78  acres  and  another  field  of  65  acres  from  a  farm  of  216  acres? 

2.  How  many  fields  of  72  acres  each  can  I  sell  from  a  farm 
containing  186  acres,  and  how  many  acres  will  remain  unsold? 

SOLUTIONS 

(1)  (2) 

78  216  72)186(2  fields 

_65  143  144 

143  sold          73  acres  unsold  42  acres  unsold 


34  REVIEW   EXAMPLES 

EXPLANATION.  In  Ex.  1,  the  remainder  is  obtained  by  adding  the  un- 
equal subtrahends  (78  +  65)  to  find  the  complete  subtrahend  (143)  and  sub- 
tracting it  from  the  given  total  of  all  the  unequal  fields  or  minuend  (216), 
obtaining  73  acres  as  the  remaining  unequal  field  [Summary,  3,  60]. 

In  Ex.  2,  multiply  the  equal  subtrahends  (72  acres)  by  the  obtained  num- 
ber of  such  subtrahends  (2)  or  quotient  [from  the  Latin,  quot,  how  many]  to 
find  the  complete  subtrahend  (144  acres);  and  subtract  this  product  or 
total  of  equal  subtrahends  (144)  from  the  dividend  or  total  of  equal  sub- 
trahends plus  a  possible  remainder  (186),  obtaining  42  acres  as  the 
remainder  [48,  k,  m,  ri\. 

SUMMARY.  1.  Employ  subtraction  to  find  the  remainder,  if  any, 
after  diminishing  a  given  total  (minuend)  by  its  known  unequal  com- 
ponent (subtrahend).  2.  Employ  division  to  find  how  many  times 
(quotient)  a  given  total  (dividend)  can  be  diminished  by  a  known 
equal  component  (divisor),  and  what  will  be  the  final  remainder,  if 
any. 

REVIEW    EXAMPLES 

Li  54.   1.  What  is  the  cost  of  853  yards  of  cloth  at  $1.85  per 
yard? 

2.  If  42  barrels  of  flour  cost  $273,  what  will  be  the  cost 
of  78  barrels  of  the  same  kind  of  flour? 

3.  What  is  the  total  cost  of  28  pounds  of  sugar  at  6  cents 
per  pound,  and  17  pounds  of  coffee  at  23  cents  per  pound? 

A  i  4.  A  man  bought  278  bushels  of  wheat  at  $1.07  per 
bushel,  327  bushels  of  oats  at  56  cents  per  bushel,  and  129  bushels 
of  rye  at  95  cents  per  bushel.  What  was  the  total  cost? 

5.  A  man  who  had  an  estate  of  $52356  willed  $25378 
to  his  wife,  $9678  to  his  son,  $12368  to  his  daughter,  and  the 
remainder  to  an  asylum.  How  much  of  the  estate  did  the 
asylum  receive  at  his  death? 

£,  4  6.  A  dealer  bought  285  tons  of  coal  at  $6.25  per  ton, 
and  sold  it  for  $1995.  What  was  his  profit? 

7.  A  speculator  bought  two  adjoining  tracts  of  land,  the 
first  containing  78  acres  at  $42  per  acre  and  the  second  contain- 
ing 94  acres  at  $35  per  acre,  expended  $2537  in  erecting  a  barn, 
$300  for  other  improvements,  and  then  sold  both  tracts  for  $12300. 
What  was  his  profit? 


REVIEW   EXAMPLES  35 

V)  '8.   A  merchant  bought  435  barrels  of  flour  at  $7.15  per 
barrel  and  sold  it  for  $3175.50.     What  was  his  gain  per  barrel? 

9.  If  I  buy  a  house  for  $6500,  pay  $3200  on  the  day  of 
purchase  and  $2758  on  a  subsequent  day,  how  much  shall  I  still 
owe? 

10.  If  I  buy  28  yards  of  muslin  at  12  cents  per  yard,  30 
yards  of  linen  at  64  cents  per  yard,  15  yards  of  cloth  at  $1.37  per 
yard,  and  give  the  seller  a  fifty-dollar  bank  note,  how  much 
change  should  I  receive? 

•4,  *11.  A  dealer  bought  75  barrels  of  potatoes  and  863 
bushes  of  corn  for  $774.34.  If  the  price  of  the  potatoes  was 
$2.50  per  barrel,  what  was  the  price  per  bushel  of  the  corn? 

12.  I  originally  contracted  a   debt   of  $3468.75.     How 
much  do  I  now  owe,  having  paid   $1268.75   at  one  time   and 
$875.36  at  another  time? 

13.  A  firm's  sales  were  as  follows:    Monday,  $375.82; 
Tuesday,    $468.74;    Wednesday,    $416.25;    Thursday,    $396.75; 
Friday,  $298.16;    and  Saturday,  $547.60.     What  were  its  average 
daily  sales? 

EXPLANATION.  To  average  or  equalize  two  or  more  unequal  quantities 
means  to  change  them  to  the  same  number  of  equal  quantities.  Hence,  the 
sum  of  these  six  unequal  quantities  when  found  should  be  regarded  as  the 
total  of  six  equal  quantities,  or  product  [Note  1,  c,  25],  and  6  should  be 
regarded  as  the  number  of  such  equal  quantities,  or  multiplier  factor  [Note 
1,  6,  25].  Hence,  divide  the  product  by  its  multiplier  factor  to  obtain  its 
multiplicand  factor,  or  one  of  the  six  equal  daily  sales. 

14.  There  were  five  boys  whose  ages  were  8,  10,  12,  14 
and  16  years,  respectively.     What  was  the  average  age  of  these 
boys? 

15.  The  temperature  of  a  certain  locality  was  75°  Fahren- 
heit for  October,  64°  for  November,  and  56°  for  December.    What 
was  the  average  temperature  for  these  three  months? 

yQ416.  A  grocer  mixed  170  pounds  coffee  at  12  cents  per 
pound,  105  pounds  coffee  at  15  cents  per  pound,  and  84  pounds 
coffee  at  18  cents  per  pound.  What  was  the  average  price  per 
pound  of  the  mixture? 

17.    If  it  is  estimated  that  the  labor  of  7  men  is  necessary 


36  REVIEW    EXAMPLES 

to  complete  a  certain  contract  in  6  days,  in  how  many  days  can 
one  man  complete  it? 

EXPLANATION.  6  days  is  a  multiplicand,  as  it  expresses  the  equal  labor 
estimated  to  be  performed  by  each  man;  9  is  a  multiplier,  as  it  expresses  the 
considered  number  of  equal  men.  Hence,  multiply  these  two  factors  to 
find  the  product  which  will  express  the  total  days'  labor  of  one  man  which  is 
performed  by  all  9  of  the  men.  Problems  of  this  character  are  solved  upon 
the  supposition  that  all  laborers  possess  equal  strength,  equal  skill,  and  equal 
industry. 

^1&.  If  35  men  can  complete  a  given  amount  of  work  in  13 
days,  in  how  many  days  can  one  man,  working  alone,  complete 
the  same  work? 

19.  If  one  man  working  alone  can  perform  a  certain  task 
in  24  days,  how  many  days  will  be  required  for  6  men  to  complete 
the  same  task? 

20.  If  50  men  can  do  a  certain  job  of  work  in  24  days, 
how  many  days  will  be  required  for  40  men  to  perform  the  same 
job? 

^  21.  A  man  was  born  in  the  year  1840  and  died  in  the  year 
1911,  how  old  was  he  at  the  time  of  his  death? 

EXPLANATION.  1840  years  after  the  birth  of  Christ  (one  given  unequal 
component  of  a  given  sum)  -f-  how  many  years  (a  required  second  unequal 
component  of  that  given  sum)  =  1911  years  after  the  birth  of  Christ  (the 
given  sum  or  minuend)  [Note  1,  b,  21]. 

22.   George  Washington  died  in  the  year  1799,  aged  67 
years.      What  was  the  year  of  his  birth? 

J  23.  The  government  of  the  United  States  was  established 
in  1789.  How  many  years  had  intervened  between  that  event 
and  the  commencement  of  the  Civil  War  in  1861? 

24.  William  H.  Taft  was  born  in  the  year  1857,  and 
was  inaugurated  President  of  the  United  States  in  the  year  1909. 
How  old  was  he  at  the  time  of  his  inauguration? 

25.  A,  B,  C  and  D  formed  a  copartnership  with  a  joint 
capital  of  $55000.     If  B  invested  twice  as  much  as  A;   C,  three 
times  as  much  as  A;  and  D,  as  much  as  B  and  C  together;   how 
much  did  each  partner  invest? 


REVIEW   EXAMPLES  37 

EXPLANATION.  Twice  (or  2  times)  is  the  given  multiplier  of  A's  invest- 
ment to  produce  B's  investment;  3  is  the  given  multiplier  of  A's  investment 
to  produce  C's  investment;  and  2  +  3,  or  5,  is  the  indirectly  given  multiplier 
of  A's  investment  to  produce  D's  investment.  Therefore,  1  time  A's  invest- 
ment or  A's  investment  +  2  times  A's  investment,  or  B's  investment  -f-  3 
times  A's  investment  or  C's  investment  +  5  times  A's  investment  or  D's 
investment,  that  is,  11  times  A's  investment,  or  all  the  partners'  investments, 
must  equal  the  joint  capital  of  the  firm  ($55000)  or  joint  product. 

Hence,  divide  the  joint  product  ($55000)  by  its  joint  multiplier  (11)  to 
find  the  common  multiplicand  or  A's  investment  ($5000).  Next  multiply  the 
obtained  common  multiplicand  ($5000)  by  the  respective  multipliers  for  B's 
investment  (2),  for  C's  investment  (3)  and  for  D's  investment  (5),  to  find 
their  respective  investment  products. 

NOTE.  If  the  sum  of  two  or  more  products  is  given,  its  multiplicand 
factor  will  be  the  primary  multiplicand  (from  which  all  the  other  mentioned 
multiplicands  have  been  successively  derived) ;  and  its  multiplier  factor  will 
be  the  sum  of  the  expressed  multipliers  +  1,  first  changing  the  multipliers 
which  have  been  expressed  upon  derivative  multiplicands,  to  equivalent 
multipliers  upon  the  primary  multiplicand  [Prin.  1,  18]. 

26.  A  and  B  together  have  $260,  B  having  three  times  as 
much  as  A.     How  much  has  each? 

27.  A,  B,  and  C  have  $520,  B  having  four  times  as  much 
as  A,  and  C  twice  as  much  as  B.     How  much  has  each? 

j  28.  A  man  bought  two  houses  for  $6275,  paying  $275 
more  for  one  than  for  the  other.  How  much  did  he  pay  for 
each? 

EXPLANATION.  $6275  is  the  given  sum  (minuend)  of  two  equal  quan- 
tities, and  $275  an  unequal  quantity  [Note  1,  6,  21];  that  is,  $6275  is  $275 
more  than  the  sum  of  two  equal  quantities,  each  denoting  the  cost  of  the 
cheaper  house.  Hence,  diminish  the  minuend  ($6275)  by  its  given  unequal 
quantity  ($275)  to  find  the  sum  of  its  two  remaining  equal  quantities  ($6000). 
As  $6000  is  the  total  of  two  equal  quantities,  or  product,  divide  it  by  its 
multiplier  factor  2  (the  number  of  equal  quantities),  obtaining  its  multipli- 
cand factor,  or  the  cost  of  the  cheaper  house  ($3000),  etc. 

29.  A  and  B  formed  a  copartnership  with  a  capital  of 
$13168,  A  investing  $568  more  than  B.     How  much  did  each 
invest? 

30.  The  firm  of  A  and  B  has  a  capital  of  $12387,  A's 
investment  being  $2387  less  than  B's.     What  is  each  partner's 
investment? 


38  REVIEW   EXAMPLES 

31.  A,  B,  and   C  together  own  property  worth  $17000, 
B's  property  being  worth  $1200  more  than  A's,  and  C's  $400 
less  than  B's.     What  is  the  worth  of  the  property  of  each? 

32.  A  man  bequeathed   $37025   to  his   wife,  son,   and 
daughter;   his  son  receiving  $625   less  than  his  wife,  and   his 
daughter  $925  more  than  his  son.     How  much  did  each  receive? 

33.  Two  ships  leave  a  certain  harbor  at  the  same  time; 
one  going  east  at  the  rate  of  350  miles  per  day,  and  the  other 
going  west  at  the  rate  of  275  miles  per  day.     How  far  apart  will 
they  be  at  the  end  of  the  ninth  day? 

34.  At  an  election  for  mayor  of  a  certain  city,  the  total 
number  of  votes  cast  for  three  candidates  was  52300.     The  first 
candidate  received  34500  votes,  and  the  second  1828  votes  less 
than  the   third.      How  many  votes   did   the    third    candidate 
receive? 

.  35.  A  man  made  five  partial  payments  on  a  debt  of 
$1565.78,  and  then  found  that  he  still  owed  $237.62.  If  his  first 
payment  was  $128.37,  his  second  $75.40,  his  third  $216.15,  and 
his  fourth  $375.28,  what  was  the  amount  of  his  fifth  payment? 

,,  36.  Two  steamers  start  from  the  same  place  at  the  same 
time  to  go  in  the  same  direction.  Four  days  after  starting  the 
faster  steamer  was  140  miles  ahead  of  the  slower.  If  the  faster 
steamer's  average  speed  was  375  miles  per  day,  how  far  was  the 
slower  steamer  from  the  starting  point? 

37.  Two  pedestrians  lived  75  miles  apart,  and  started 
from  their  respective  homes  at  the  same  time  to  approach  each 
other,  one  walking  at  the  rate  of  four  miles  per  hour,  and  the 
other  at  the  rate  of  three  miles  per  hour.  How  far  apart  were 
they  after  walking  eight  hours? 

,  38.  A  farmer  exchanged  a  tract  of  land  containing  275 
acres,  and  valued  at  $45  per  acre  for  another  tract  containing  495 
acres.  What  was  the  estimated  value  per  acre  of  the  other 
tract? 

39.  A  certain  library  contains  43  book  cases,  each  case 
containing  12  shelves,  and  each  shelf  78  books.  How  many 
books  does  the  library  contain? 

•   40.   A  merchant  examined  the  contents  of  his  cash  drawer 


FACTORING  39 

and  found  that  it  contained  $48.  Of  this  sum  he  found  $30  to 
be  in  notes,  and  the  remainder  consisted  of  an  equal  number  of 
50-cent  coins,  25-cent  coins,  dimes  and  nickels.  How  many  coins 
did  he  find  in  the  cash  drawer? 


FACTORING 

55.  Factoring  is  that  process  by  which  a  number  is  separated 
into  two  or  more  integers  [Note  1,  e,  3]  which,  when  multiplied, 
will  produce  that  number. 

NOTE  1.  Each  of  the  integers  into  which  a  number  is  separated  as  above 
described,  is  called  a  factor  of  that  number. 

NOTE  2.  When  a  number  cannot  be  separated  into  integral  factors  other 
than  itself  or  1,  it  is  called  a  prime  number;  and  when  it  can  be  so  separated, 
it  is  called  a  composite  number. 

NOTE  3.  The  factors  of  a  number  are  also  distinguished  as  prime  factors 
when  they  are  not  separable  into  minor  integral  factors;  and  as  composite 
factors  when  they  are  so  separable. 

56.  An  exact  divisor  of  a  number  is  one  which  will  divide  it 
without  a  remainder. 

NOTE.  Reversely,  one  number  is  said  to  be  divisible  by  another  when  it 
contains  that  other  a  certain  number  of  times  without  a  remainder. 

57.  Any  number  is  divisible  by  itself  and  1;    and  it  will  be 
divisible: 

(a)  By  2,  if  it  ends  with  a  cipher,  or  with  a  figure  which  is 
divisible  by  2. 

Thus,  480  and  7818  are  divisible  by  2;  the  former  because  it  ends  with  a 
cipher,  the  latter  because  it  ends  with  8  which  is  divisible  by  2. 

(6)  By  3,  if  the  sum  of  all  its  figures  is  divisible  by  3. 

Thus,  4791  is  divisible  by  3  because  4  +  7  +  9  +  1,  or  21,  is  divisible 
by  3. 

(c)  By  4,  if  it  ends  with  two  ciphers,  or  with  two  figures  which 
express  a  number  that  is  divisible  by  4. 

Thus,  34700  and  72324  are  divisible  by  4;  the  former  because  it  ends  with 
two  ciphers,  the  latter  because  it  ends  with  two  figures  (24)  which  express 
a  number  that  is  divisible  by  4. 


40  FACTORING 

(d)  By  5,  if  it  ends  either  with  a  cipher  or  with  5. 

Thus,  720  and  1465  are  divisible  by  5;  the  former  because  it  ends  with 
a  cipher,  and  the  latter  because  it  ends  with  5. 

(e)  By  6,  if  it  ends  with  a  figure  which  is  divisible  by  2  (one 
factor  of  6),  and  the  sum  of  all  its  figures  is  divisible  by  3  (the  other 
factor  of  6). 

Thus,  1368  is  divisible  by  6  because  it  ends  with  a  figure  (8)  which  is 
divisible  by  2;  and  1  +  3  +  6  +  8,  or  18,  is  divisible  by  3. 

(/)  By  8,  if  it  ends  with  three  ciphers,  or  with  three  figures 
which  express  a  number  that  is  divisible  by  8. 

Thus,  276000  and  18376  are  divisible  by  8;  the  former  because  it  ends  with 
three  ciphers,  and  the  latter  because  it  ends  with  three  figures  (376)  which 
express  a  number  that  is  divisible  by  8. 

(g)  By  9,  if  the  sum  of  its  figures  is  divisible  by  9. 

Thus,  32472  is  divisible  by  9  because  3  +  2  -f  4  +  7  +  2,  or  18,  is  divisible 
by  9. 

(h)  By  10,  if  it  ends  with  one  or  more  ciphers. 
Thus,  6730  is  divisible  by  10  because  it  ends  with  a  cipher. 

(i)  By  12,  if  it  ends  with  two  figures  which  express  a  number 
that  is  divisible  by  4  (one  factor  of  12),  and  the  sum  of  all  its 
figures  is  divisible  by  3  (the  other  factor  of  12). 

Thus,  71928  is  divisible  by  12  because  it  ends  with  two  figures  (28)  which 
express  a  number  that  is  divisible  by  4,  and  7  +  1  +  9  -f  2  +  8,  or  27,  is 
divisible  by  3. 

ILLUSTRATIVE   EXAMPLE 
58.   What  are  the  prime  factors  of  630? 

SOLUTION  EXPLANATION.     As  630  ends  with  a  cipher,  divide  it  by  its 

prime  factor  5  [57,  d\,  obtaining  126  as  the  unfactored  part  of 

5)630         630.     As  126  ends  with  6  which  is  divisible  by  2,  the  entire  un- 

2)126         factored  part  is  so  divisible  [57,  a],  obtaining  63  as  the  remain- 

o\™          ing  unfactored  part  of  the  original  number.     As  the  sum  of  the 

figures  of  63  (6  +  3),  or  9,  is  divisible  by  3,  the  entire  unfactored 

3)21          part  (63)  is  so  divisible,  obtaining  21  as  the  remaining  unfactored 

"7         part  [57,  6].     As  the  sum  of  the  figures  of  21  (2  +  1),  or  3,  is 

divisible  by  3,  the  entire  unfactored  part  (21)  is  so  divisible, 

obtaining  7,  a  prime  number.     Therefore,  the  several  prime  divisors  and 


LEAST    COMMON    DIVIDEND  41 

the  final  prime  quotient  7,  that  is,  5,  2,  3,  3,  7,  are  the  required  prime  factors 
of  630. 

PROOF.  The  continued  product  of  all  its  prime  factors  should  equal  the 
factored  number.  Thus  5X2X3X3X7  should  equal  630. 

RULE.  1.  Divide  the  given  composite  number  by  any  prime 
number  which  will  exactly  divide  it.  2.  Divide  the  resulting  quo- 
tient by  any  prime  number  which  will  exactly  divide  it;  and  so 
continue  until  a  quotient  is  obtained  which  is  a  prime  number.  3. 
The  several  divisors  and  the  final  quotient  will  be  the  prime  factors 
of  the  given  composite  number. 

EXAMPLES  FOR  PRACTISE 

Find  the  prime  factors  of 

1.  54     4.  360     7.  1260     10.  27000     13.  16632 

2.  36     6.  252     8.  5292     11.  17640     14.  45000 

3.  84     6.  945     9.  3375     12.  28512     15.  16380 


LEAST  COMMON  DIVIDEND 

59.  A  dividend  of  a  number  is  one  that  contains  it  an  exact 
number  of  times. 

Thus,   63  is  a  dividend  of  7  because  63  contains  7  exactly  nine  times. 

60.  A  common   dividend  of  two  or  more  numbers  is  one  that 
contains  each  of  them  an  exact  number  of  times. 

Thus,  24  is  a  common  dividend  of  3,  6  and  8,  because  it  contains  3 
exactly  eight  times,  6  exactly  four  times,  and  8  exactly  three  times. 

61.  The  least   common    dividend    of    two  or  more   numbers 
is  the  least  number  which  contains  each  of  them  an  exact  number 
of  times. 

62.  General  principles.     1.  A  dividend  of  a  number  must  in- 
clude all  its  prime  factors,  and  may  include  other  factors.     2.  A 
common  dividend  of  two  or  more  numbers  must  include  all  the  prime 
factors  of  each  of  those  numbers,  and  may  include  other  factors. 
3.    The  least  common  dividend  of  two  or  more  numbers  must  include 
all  the  prime  factors  of  those  numbers,  and  no  other  factor. 


42  LEAST    COMMON    DIVIDEND 

ILLUSTRATIVE  EXAMPLE 
63.    Find  the  least  common  dividend  of  12,  18,  and  20. 

SOLUTION  EXPLANATION.     2    is    a  common    prime 

factor     of     each     of     the     given     numbers, 
—  18  —  20  therefore,  the  least  common  dividend  must 


2)6  —     9—10  contain  the  factor  2  at  least  once. 

3)3~I       Q~      5  Of  the  unfactored  part  of  12  (6),  of  18  (9) 

and  of  20  (10),  2  is  a  common  prime  factor 

1—3—5  of  6  and  10.     Therefore  the  least  common 

dividend  must  contain  the  factor  2  at  least 

2X2X3X3X5=  180     twice  to  contain  12  and  20  which  include  this 

factor  twice. 

Of  the  yet  unfactored  part  of  12  (3),  of  18  (9)  and  of  20  (5),  3  is  a 
common  prime  factor  of  3  and  9.  Therefore  the  least  common  dividend 
must  contain  the  factor  3  at  least  once  in  order  to  contain  12  and  18  which 
include  this  factor  at  least  once. 

The  least  common  dividend  must  therefore  not  only  contain  the  prime 
factors  already  found  (2,  2,  3),  but  also  the  yet  unfactored  part  of  18  (3)  or  it 
cannot  contain  18,  and  the  yet  unfactored  part  of  20  (5)  or  it  cannot  contain 
20.  Therefore  the  least  common  dividend  of  12,  18  and  20  must  contain  these 
necessary  factors  (2,  2,  3,  3  and  5)  and  no  other  factor,  and  the  continued  product 
of  these  factors  is  180. 

NOTE  1.  When  no  two  of  the  given  numbers  have  a  common  prime 
factor,  the  least  common  dividend  of  those  numbers  must  be  their  continued 
product. 

NOTE  2.  When  it  is  seen  that  all  of  the  given  numbers  have  a  common 
composite  factor  [Notes  2  and  3,  65],  it  will  save  unnecessary  labor  to  divide 
at  once  by  the  composite  factor. 

NOTE  3.  When  one  of  the  given  numbers  of  which  it  is  required  to  find 
the  least  common  dividend  is  found  to  be  an  exact  divisor  of  another  of 
the  given  numbers,  it  may  be  omitted  from  the  solution,  as  all  of  its 
factors  must  then  be  included  in  that  other  number. 

EXAMPLES  FOR  PRACTISE 
Find  the  least  common  dividend  of 

1.  15,  18,  24,  and  36  7.   360,  450,  540,  and  630 

2.  21,  49,  56,  and  63  8.    150,  500,  600,  and  900 

3.  18,  27,  45,  and  72  9.    132,  165,  231,  and  297 

4.  24,  36,  48,  and  64  10.   5,  15,  25,  30,  45  [Note  3] 

5.  27,  36,  54,  and  81  11.    12,  36,  48,  72,  and  96 

6.  30,  45,  75,  and  90  12.   2,    3,    7,   11,    13  [Note  l] 


GREATEST    COMMON    DIVISOR  43 

GREATEST  COMMON  DIVISOR 

64.  A   common  divisor  of  two  or  more  numbers  is  an  exact 
divisor  of  each  of  them. 

Thus,  7  is  a  common  divisor  of  21,  35  and  56,  as  7  is  an  exact  divisor,  or 
common  factor,  of  each  of  them. 

65.  The  greatest    common    divisor  of  two  or  more  numbers  is 
the  greatest  exact  divisor  of  each  of  them. 

Thus,  while  2  is  a  common  divisor  of  24,  40  and  56,  it  is  not  the  greatest 
common  divisor;  and  while  4  is  a  greater  common  divisor  of  these  numbers 
than  2,  it  is  also  not  the  greatest  common  divisor.  The  greatest  divisor  com- 
mon to  these  numbers  is  8,  because  8  is  the  greatest  factor  common  to  all  of 
them. 

66.  General  principles.     1.   A  common  divisor  of  two  or  more 
numbers  must  include  in  itself  only  such  prime  factors  as  are  common 
to  those  numbers. 

Thus,  6  is  a  common  divisor  of  30  and  42,  because  all  the  prime  factors 
of  6  (3  X  2)  are  found  both  in  30  (3  X  2  X  5)  and  in  42  (3  X  2  X  7). 

2.  The  greatest  common  divisor  of  two  or  more  numbers  must  in- 
clude all  the  prime  factors  which  are  common  to  those  numbers,  and 
no  other  factor. 

Thus,  9  is  the  greatest  common  divisor  of  45  and  63  because  all  the 
prime  factors  which  are  common  to  45  (3  X  3  X  5)  and  to  63  (3  X  3  X  7) 
are  also  found  in  9  (3  X  3).  A  greater  composite  number  than  9  cannot  be 
a  common  divisor  of  45  and  63  because  it  will  include  a  factor  which  is  not 
common  to  45  and  63;  and  a  less  number  than  9,  say  3,  which  contains  one 
factor  common  to  45  and  63  (3  X  1),  and  therefore  a  common  divisor,  would 
not  contain  all  the  common  factors  of  those  numbers  and  therefore  could 
not  be  their  greatest  common  divisor. 

3.  The  greatest  common  divisor  of  two  numbers  must  also  be  a 
divisor  (a)  of  their  sum,  (b)  of  their  difference,  (c)  of  any  number  of 
times  their  sum,  (d)  or  of  any  number  of  times  their  difference. 

4.  The  greatest  common  divisor  of  two  numbers  must  also  be  (e) 
a  divisor  of  any  number  of  times  each  of  them;  or  (f)  of  any  number 
of  times  one  of  them  and  a  different  number  of  times  the  other;  or  (g) 
of  the  difference  between  the  greater  number  and  any  number  of  times 
the  kss  number  that  can  be  subtracted  from  the  greater  number. 


44  GREATEST    COMMON    DIVISOR 

5.  The  greatest  common  divisor  of  more  than  two  numbers  can- 
not be  greater  than  the  greatest  common  divisor  of  the  two  which 
express  the  fewest  units. 

ILLUSTRATIVE  EXAMPLE 
67.   Find  the  greatest  common  divisor  of  531  and  767. 

EXPLANATION.     The  greatest  divisor  of  531 

SOLUTION  could  not  also  be  the  greatest  divisor  of  767  except 

ro-i  \T«T/i  it  be  also  a  divisor  of  their  difference  which  is  236 

531)76'  [Prin.  3,  b,  66]. 

Therefore,  the  G.  C.  D.  of  531  and  767  must 

236)531(2  also  be  the  G.  C.  D.  of  236,  531  and  767;  and  this 

472  G-  C.  D.  cannot  be  greater  than  the  G.  C.  D.  of 

—  236  and  531    [Prin.   5,   66].      Now  the  greatest 

divisor  of  236   could   not  also  be  the    greatest 

236          divisor  of  531  except  it  be  also  a  divisor  of  the 

difference  between  531   and  twice  236,   or   472, 

which  is  59  [Prin.  4,  g,  66]. 

Therefore,  the  G.  C.  D.  of  236,  531  and  767  must  also  be  the  G. 
C.  D.  of  59,  236,  531  and  767,  and  this  G.  C.  D.  cannot  be  greater  than  the 
G.  C.  D.  of  59  and  236  [Prin.  5,  66].  As  59  will  divide  itself  and  as  it  is  also 
found  to  be  an  exact  divisor  of  236,  59  is  the  G.  C.  D.  of  59  and  236,  and  as 
236  is  the  difference  between  531  and  767,  59  must  also  be  the  G.  C.  D.  of  531 
and  767  [Prin.  3,  6,  66]. 

RULE.  1.  Divide  the  greater  number  by  the  less;  and  if  there  be 
no  remainder,  the  less  number  will  be  the  greatest  common  divisor. 

2.  Divide  the  last  divisor  by  the  remainder  if  there  should  be  a 
remainder,  and  if  the  division  be  exact,  that  remainder  will  be  the 
greatest  common  divisor. 

3.  //  the  second  division  be  inexact,  continue  to  divide  the  last 
divisor  by  the  last  remainder  until  there  shall  be  no  remainder.     The 
last  divisor  will  be  the  greatest  common  divisor. 

NOTE  1.  If,  at  any  time,  1  is  obtained  as  a  remainder,  it  will  demonstrate 
that  the  numbers  are  prime  to  each  other. 

NOTE  2.  To  find  the  G.  C.  D.  of  more  than  two  numbers,  first  find  the 
G.  C.  D.  of  two  of  them,  preferably  of  the  two  which  express  the  fewest  units; 
then  find  the  G.  C.  D.  of  this  obtained  G.  C.  D.  and  one  of  the  remaining 
numbers;  and  thus  continue  until  all  the  given  numbers  have  been  considered. 
The  last  obtained  G.  C.  D.  will  be  the  G.  C.  D.  of  all  the  numbers. 

Note  3.  If  the  numbers  are  not  inconveniently  great,  their  greatest 
common  divisor  may  also  be  obtained  by  finding  the  continued  product  of 
such  prime  factors  as  are  common  to  all  of  them. 


CANCELATION  45 

EXAMPLES  FOR  PRACTISE 

Find  the  greatest  common  divisor  of 

1.  185  and  259  5.  539  and  637  9.  16767  and  29889 

2.  315  and  405  6.  10625  and  14375  10.  14406  and  26411 

3.  204  and  612  7.  1344  and  4416  11.  12096,  19008,  22464 

4.  875  and  1375  8.  1215  and  2673  12.  704,  448,  880 

CANCELATION 

68.  Cane  elation  is  the  process  of  diminishing  the  mechanical 
labor  of  division  by  casting  out  factors  which  are  common  to  both 
divisor  and  dividend. 

Thus,  in  dividing  168  by  56,  if  it  be  noticed  that  8  is  a  common  factor 
of  both  divisor  and  dividend,  the  labor  of  division  will  be  diminished  by 
mentally  dividing  one-eighth  of  168,  or  21,  by  one-eighth  of  56,  or  7,  obtaining 
3  as  the  quotient  [Prin.  5,  34]. 

69.  General  principles.      1.    To  cancel  any  factor  of  a  number 
has  the  same  effect  as  dividing  that  number  by  the  canceled  factor. 

2.  To  cancel  the  same  factor,  or  the  same  combination  of  factors, 
or  equivalent  combinations  of  factors,  from  both  divisor  and  dividend, 
does  not  affect  the  value  of  the  quotient  [Prin.  5,  34]. 

ILLUSTRATIVE  EXAMPLE 

The  factors  of  a  dividend  are  210,  16  and  18,  and  of  its  divisor 
are  27,  8  and  7.  What  is  the  quotient? 

SOLUTION  EXPLANATION.     Instead  of  finding  the  continued 

}0  product  of  210,   16  and   18,  which  is  60480,  and 

00          2          o  dividing  this  result  by  the  continued  product  of 

TT        ffl  27  X  8  X  7,  which  is  1512,  obtaining  40,  the  me- 

/P  X  /P_  .     chanical  labor  of  the  solution   can  be  materially 


$7  X     $  X     /7  diminished    by  cancelation,  or    continued  division, 

£  [37],  as  follows: 

Place  the  factors  of  the  dividend  above  a  hori- 

zontal line  and  the  factors  of  the  divisor  below  that  line.  Then  cancel  the 
common  factor  9  from  27  in  the  divisor  and  from  18  in  the  dividend,  writ- 
ing underneath  27  its  uncanceled  factor  3,  and  above  18  its  uncanceled 
factor  2.  Then  cancel  the  common  factor  8  from  8  in  the  divisor  and  from 
16  in  the  dividend,  and  over  16  place  its  uncanceled  factor  2.  Then  cancel 
the  common  factor  7  from  7  in  the  divisor  and  from  210  in  the  dividend,  and 


46  COMMON    FRACTIONS 

over  210  place  its  uncanceled  factor  30.  Then  cancel  the  yet  uncanceled 
factor  3  from  3  in  the  divisor  and  from  30  in  the  dividend,  and  place  over 
30  its  uncanceled  factor  10.  The  product  of  the  uncanceled  factors  in  the 
dividend  (10  X  2  X  2  =  40)  will  be  the  required  quotient. 

NOTE  1.  If,  after  the  cancelation  has  been  completed,  any  uncan- 
celed factors  are  still  found  in  the  divisor,  the  product  of  the  uncanceled 
factors  of  the  dividend  should  be  divided  by  the  product  of  the  uncanceled 
factors  of  the  divisor,  to  obtain  the  quotient. 

NOTE  2.  If  the  product  of  two  or  more  factors  on  one  side  of  the  hori- 
zontal line  is  seen  to  be  equal  to  a  single  factor  on  the  other  side  of  that  line, 
they  should  be  canceled  in  one  operation. 

NOTE  3.  If  1  is  obtained  as  a  result  of  cancelation,  it  should  be  ex- 
pressed in  the  dividend,  but  it  may  be  omitted  from  the  divisor. 

EXAMPLES  FOR  PRACTISE 

1.  Divide  73  X  27  X  16  by  18  X  4. 

2.  Divide  145  X  36  X  12  by  30  X  18  X  4. 

3.  Divide  648  X  75  X  32  by  54  X  45  X  8. 

4.  Divide  180  X  63  X  24  by  48  X  18  X  9. 

5.  How  many  bushels  of   wheat  worth  $1.12    per  bushel 
should  a  farmer  deliver  at  a  mill  to  receive  3  barrels  of  flour  worth 
$5.60  per  barrel? 

6.  A  speculator  offered  a  tract  of  land  containing  375  acres 
in  exchange  for  250  acres  of  land  worth  $75  per  acre.     At  what 
price  per  acre  did  the  speculator  value  his  land? 

COMMON  FRACTIONS 

70.  INTRODUCTION.  Hitherto,  all  the  operations  which  have 
been  described  and  illustrated  have  been  performed  upon  inte- 
gral units  [Note  1,  e,  3].  It  is  frequently  necessary,  however,  to 
operate  upon  a  part  of  an  integral  unit,  as  upon  the  half  of  one 
integral  unit  (i),  the  third  of  one  integral  unit  (J),  etc.;  or  upon 
a  part  of  a  collection  of  integral  units,  as  upon  a  third  of  two  inte- 
gral units  (|),  an  eighth  of  five  integral  units  (f ),  etc.  The  general 
principles  already  learned  are  unlimited  in  their  application  to 
all  numbers,  whether  integral  or  fractional. 

Any  perplexity  attending  the  study  of  this  new  fractional  form 
of  numerical  expression  will  disappear  if  every  fraction  be  regarded 
as  an  unexecuted  division  of  which  its  numerator  is  the  dividend, 
its  denominator  is  the  divisor,  and  the  value  of  the  entire  frac- 
tional expression  is  the  quotient.  Thus  regarded,  the  general 


COMMON    FRACTIONS  47 

principles  relating  to  division  [34],  so  far  as  they  apply  to  divi- 
dends, will  also  apply  to  numerators  [72];  so  far  as  they  apply 
to  divisors,  will  also  apply  to  denominators  [73];  and  so  far  as 
they  apply  to  quotients,  will  also  apply  to  the  entire  fractional 
expression  [71].  All  reductions  of  fractions  [82],  therefore,  are 
applications  of  the  general  principles  of  34. 

71.  A  fraction  expresses  the  result  of  dividing  one  or  more 
units  into  a  given  number  of  equal  parts. 

72.  The  numerator  is  that  term  of  a  fraction  which  is  written 
above  a  horizontal  line,  and  which  denotes  how  many  of  the  equal 
parts  are  expressed. 

Thus,  $f  expresses  5  fractional  units,  the  value  of  each  of  which  is  one- 
eighth  of  a  dollar.  Num(b)er-ator  means  that  which  numbers. 

73.  The   denominator   is   that   term   of   a   fraction  which   is 
written  below  a  horizontal  line,  and  which  denotes  the  value  of 
each  fractional  unit  expressed  by  the  numerator. 

Thus,  in  the  fraction  $f,  the  numerator  (4)  denotes  the  number  of  frac- 
tional units;  and  the  denominator  (5)  names  the  value  of  each  fractional  unit 
as  being  one-fifth  of  a  dollar.  Denominator  means  that  which  denominates  or 
names. 

74.  The  terms  of  a  fraction  are  its  numerator  and  denomina- 
tor, considered  separately. 

Thus,  in  |,  the  numerator  7  is  one  term  and  the  denominator  8  is  the 
other.  Considered  jointly,  they  constitute  the  fraction  f . 

75.  A  common  fraction  is  one  which  expresses  the  division  of 
one  or  more  units  into  any  number  of  equal  parts  other  than 
such  as  are  represented  by  1  followed  by  one  or  more  ciphers. 

Thus,  |,  T5*,  *ib  are  common  fractions;  but  T%,  T&T,  Tolhr  are  not. 

76.  A  proper  fraction  is  one  in  which  the  numerator  is  less 
than  the  denominator. 

A  proper  fraction  Is  so-called  because  a  fraction  properly  means  part 
of  a  whole;  and  if  the  numerator  of  a  fraction  should  exceed  its  denominator, 
it  would  express  more  than  a  whole  and  could  not  properly  be  called  a  fraction, 
though  for  convenience  it  may  be  so  indicated. 

77.  An  improper  fraction  is  one  in  which  the  numerator  either 
equals  or  exceeds  its  denominator. 


48  COMMON    FRACTIONS 

Thus,  |,  1  are  proper  fractions  because  the  value  of  each  is  less  than  1 
integral  unit;  and  |,  f  are  improper  fractions,  the  former  because  its  value  is 
exactly  equal  to  1  integral  unit,  and  the  latter  because  its  value  is  greater 
than  1  integral  unit. 

NOTE.  Any  integer  may  be  expressed  in  the  form  of  an  improper  fraction 
by  writing  1  under  it  for  a  denominator.  Thus,  5  =  f;  19  =  ±T. 

78.  A  mixed   number   is   an   expression  which   is   composed 
partly  of  an  integer  and  partly  of  a  fraction. 

Thus,  35 1  is  suggestively  called  a  mixed  number  because  it  is  composed 
partly  of  an  integer  (35)  and  partly  of  a  fraction  (£),  mixed  or  united  in  one 
expression  (35|). 

79.  The  value  of  a  fraction  is  the  value  expressed  by  con- 
summating the  unexecuted  division  of  its  numerator  by  its  de- 
nominator. 

Hence,  the  nearer  a  numerator  of  a  proper  fraction  approximates  to  its 
denominator,  the  greater  its  value;  because  any  increase  in  a  dividend  (num- 
erator), the  divisor  (denominator)  remaining  the  same,  must  necessarily 
increase  the  value  of  the  quotient.  Prin.  4,  34,  relating  to  an  increase  of  the 
dividend  (numerator)  by  multiplication  must,  though  in  a  different  manner, 
apply  also  to  an  increase  by  addition. 

80.  (a)  The  unit  of  a  fraction  is  one  of  the  integral  units  of 
which  the  fraction  expresses  a  part;   and  (6)  a  fractional  unit  is 
one  of  the  equal  parts  into  which  the  unit  of  the  fraction  is  divided. 

Thus,  in  the  expression  $f,  one  dollar  is  the  unit  of  the  fraction;  and 
one  eighth-of-a-dollar  is  the  fractional  unit. 

81.  General  principles.     1.    To  multiply  or  to  divide  the  num- 
erator by  any  number  is  equivalent  to  multiplying  or  to  dividing  the 
fraction  by  that  number  [Prin.  4,  34]. 

2.  To  multiply  the  denominator  by  any  number  is  equivalent  to 
dividing  the  fraction  by  that  number  [Prin.  6,  34]. 

3.  To  divide  the  denominator  by  any  number  is  equivalent  to 
multiplying  the  fraction  by  that  number  [Prin.  7,  34]. 

4.  To  multiply  or  to  divide  both  the  numerator  and  the  denomina- 
tor by  the  same  number  will  produce  no  change  in  the  value  of  the 
fraction  [Prin.  5,  34]. 

SUMMARY,  (a)  Any  change  in  the  numerator  by  multiplication  or  divi- 
sion, the  denominator  remaining  the  same,  will  produce  a  similar  change  in 


COMMON    FRACTIONS  49 

the  value  of  the  fraction;  (6)  Any  change  in  the  denominator  by  multiplica- 
tion or  division,  the  numerator  remaining  the  same,  will  produce  an  opposite 
change  in  the  value  of  the  fraction;  (c)  A  similar  change  in  both  numerator 
and  denominator,  by  multiplication  or  division,  will  produce  no  change  in  the 
value  of  the  fraction. 

REDUCTION    OF    FRACTIONS   TO    HIGHER   TERMS 

82.  Reduction  of  fractions  is  the  process  of  changing  their 
form  (appearance)  without  producing  any  change  in  their  value. 

Thus,  if  it  is  desired  to  change  a  fraction  in  low  terms  (possessing  a  less 
numerator  and  denominator)  to  an  equivalent  fraction  in  higher  terms  (pos- 
sessing a  greater  numerator  and  denominator),  the  name  of  the  process  is 
reduction,  and  of  the  act,  to  reduce. 

ILLUSTRATIVE  EXAMPLE 
Reduce  f  to  126ths. 

SOLUTION  EXPLANATION.     The  required  denominator  (126) 

is  18  times  the  given  denominator  7  (126  •*-  7  =  18); 

l^u  —  '    ==  1*       therefore,    the    proper    numerator    for    the    required 

«        -~  _  ..,.£       denominator  must  be  18  times  the  given  numerator 

6  (6  X  18  =  108),  to  maintain  a  value  equivalent  to 

7  X  18  =  126       that  of  f  [Prin.  4,  81]. 

RULE.  Multiply  both  the  numerator  and  the  denominator  by  such 
a  number  as  will  change  the  given  denominator  to  the  required  de- 
nominator. 

NOTE  1.  To  find  the  common  multiplier  of  both  terms  to  produce  a 
required  higher  denominator,  divide  the  required  denominator  (the  greater) 
by  the  given  denominator  (the  less). 

NOTE  2.  To  reduce  an  integer  to  a  fraction  of  a  required  denominator, 
first  change  the  integer  to  the  form  of  a  fraction  by  writing  the  denominator 
1  under  it,  or  by  conceiving  it  to  be  so  written.  •  Thus,  5  =  yj  19  =  *i  ',  etc. 

EXAMPLES  FOR  PRACTISE 

Reduce  Reduce  Reduce 

1.  £to25ths                5.   f  to216ths  9.  J  to  256ths 

2.  f  to  56ths                6.   f  to  512ths  10.  T\  to  64ths 

3.  f  to32nds               7.   }  to  16ths  11.  28  to  4ths 

4.  \  to  64ths                8.   I  to  128ths  12.  176  to  16ths 


50  COMMON    FRACTIONS 

REDUCTION    OF    FRACTIONS   TO   LOWER   TERMS 

ILLUSTRATIVE  EXAMPLE 
83.   Reduce  &6*  to  32nds. 

SOLUTION  EXPLANATION.     First  divide  the  given  denominator 

_  (544)  by  the  required  denominator  (32)  to  find  the 

**      divisor  of  the  given  denominator    (544)   which  will 

OK  _._  17  _    c      produce  the  required  denominator  (32),  obtaining  17. 

-   *  If  the  given  denominator  must  be  divided  by  17 

544  -r-  17  =  32     to  find  the  required  denominator,  the  given  numerator 

(85)  must  also  be  divided  by  17  to  find  the  corre- 

sponding numerator  (5)  of  the  required  denominator  [Prin.  4,  81]. 

RULE.  Divide  both  the  numerator  and  the  denominator  by  such 
a  common  divisor  as  will  change  the  given  denominator  to  the  required 
denominator. 

NOTE.  To  find  the  common  divisor  of  both  terms  of  a  fraction  which  will 
produce  an  equivalent  fraction  of  the  required  lower  denominator,  divide  the 
given  denominator  (the  greater)  by  the  required  denominator  (the  less). 

EXAMPLES  FOR  PRACTISE 
Reduce                      Reduce  Reduce 

1.  f  M  to  8ths         4.  iji  to  32nds         7.  J3i  to  64ths 

2.  -ffy  to  16ths       5.  jfi  to  halves        8.  ^-f*  to  an  integer  (firsts) 

3.  TW  to  4ths         6.  -Bf  to  4ths  9.  Vff-  to  an  integer 


REDUCTION  OF  FRACTIONS  TO  LOWEST  TERMS 

84.   A  fraction  is  said  to  be  reduced  to  lowest  terms  when  its 
numerator  and  denominator  are  prime  to  each  other. 

ILLUSTRATIVE  EXAMPLES 
Reduce  (1)  J|J  and  (2)  fj&  to  lowest  terms. 

SOLUTIONS 

(1)  (2) 

168  -s-  8  =  21 
384  -v-  8  =  48         G.  C.  D.  of  949  and  2336  =  73 

21  •*-  3  =  7_         949  -T-  73  =  13 

48  -s-  3  =  16  AnS'     2336  -J-  73  =  32   nS' 


COMMON    FRACTIONS  51 

EXPLANATIONS.  (1)  Divide  both  terms  of  Iff  by  any  observed  common 
factor,  say  8,  obtaining  |J  as  an  equivalent  fraction  in  lower  terms  [Prin.  4, 
81].  It  will  now  be  further  observed  that  3  is  a  common  factor  of  both  terms  of 
ft.  Hence,  divide  both  terms  of  \ \  by  3,  obtaining  TV  as  an  equivalent  frac- 
tion in  still  lower  terms.  As  the  terms  of  yV  are  prime  to  each  other,  they  are 
incapable  of  further  reduction,  and  therefore  T^  is  in  its  lowest  possible  terms. 

(2)  If  the  factor  or  factors  common  to  both  terms  are  not  easily  discover- 
able, as  in  2T&,  it  will  be  more  convenient  to  divide  both  numerator  and 
denominator  by  their  greatest  common  divisor  [67]. 

EXAMPLES  FOR  PRACTISE 
Reduce  the  following  fractions  to  lowest  terms: 

i.  jt       5.  m        9.  ««       is. 

2.  if  6.    »J  10.   IIH  14. 

3.  ||  7.   }}{  11.   Uii  15.   Hill 

4.  «  8.   HI  12.   tffi  16.   Httt 

REDUCTION    OF   FRACTIONS   TO   APPROXIMATE 
LOWEST   TERMS 

85.  INTRODUCTION.   To    approximate    means    to    approach 
very  near  without  exactly  reaching.    Approximations  are  frequently 
necessary  in  business  calculations  to  secure  serviceable  results. 
The  market  reports  of  our  daily  papers,  contain  no  fractional  quo- 
tations  other   than   halves,    quarters,    eighths,    and   sometimes, 
though  rarely,  sixteenths.     In  exceptional  occupations  which  re- 
quire other  denominators  than  the  above,  any  obtained  fractional 
results    are    approximated    to    such    exceptional    denominators. 
Hence,  if  the  final  result  of  a  calculation  contains  a  fraction  with 
an  impractical  denominator,  it  is  customary  to  reduce  it  to  a 
practical  denominator  by  approximation. 

ILLUSTRATIVE  EXAMPLES 

86.  Reduce  (1)  H  to  8ths;  (2)  f  Jf  to  4ths. 

FIRST  SOLUTION  FIRST  EXPLANATION.  If  the  division  of 

21  X  8  =  168  *ts  numerator  by  its  denominator  expresses  the 

value  of  any  given  fraction  [79],  then  to  divide 
168  -r-  56  =  3,  exactly  8  times  the  numerator  of  U  (21  X  8  =  168)  by 

the  denominator  of  fi  (168  -5-  56)  must  express 
Hence,  f  £  =  I,  exactly  a  result  (exactly  3)  which  is  exactly  8  times  the 

value  of  H  [Prin.  1,  81];  and  this  result  (3) 
divided  by  8,  or  f ,  must  be  the  exact  value  of  H  expressed  in  eighths. 


52  COMMON   FRACTIONS 

SECOND    SOLUTION  SECOND      EXPLANATION.        To 

413  X  4  =  1652  divide  4  times  the  numerator  of 

f  H  (413  X  4  =  1652)  by  the 

1652  -T-  548  =  3,  approximately  denominator  of  fit  (1652  ^  548) 

will  produce  a  result  (3,  approxi- 

Hence,  HI  =  I,  approximately  mately)  which  must  be  approxi- 
mately 4  times  the  value  of  |?f 

[Prin.   1,   81].      Hence,   this  result   (3)   divided  by  4,  or    f,   must  be  the 

approximate  value  of  f  if  when  expressed  in  fourths. 

RULE.  Multiply  the  numerator  of  the  given  fraction  by  the 
required  denominator,  and  divide  the  resulting  product  by  the  de- 
nominator of  the  given  fraction,  disregarding  any  remainder  from 
the  division  if  less  than  half  the  divisor,  or  increasing  the  quotient  by 
1  if  the  remainder  is  half  the  divisor,  or  more.  The  quotient  will  be 
the  exact  numerator  if  the  division  be  exact,  or  the  approximate  num- 
erator if  the  division  be  inexact.  Under  the  quotient  place  the  required 
denominator.  If  the  resulting  fraction  is  not  in  lowest  terms,  men- 
tally apply  Solution  1,  84. 

NOTE.  Intermediate  fractions  should  never  be  approximated  as  the 
least  departure  from  strict  accuracy  near  the  beginning  of  a  continued 
operation  may  terminate  in  a  considerable  error  when  the  final  result  is 
reached. 

EXAMPLES   FOR  PRACTISE 
Reduce  the  following  to  lowest  terms : 

1.  iff,  not  higher  than  8ths  6.   fj£i,  not  higher  than  16ths 

2.  ^T£,  not  higher  than  4ths  6.   if  ft,  not  higher  than  8ths 

3.  Iff,  not  higher  than  16ths         7.   IHI,  not  higher  than  32nds 

4.  *§!,  not  higher  than  8ths  8.    ||f-J-,  not  higher  than  4ths. 

REDUCTION    OF    INTEGRAL    OR    MIXED    NUMBERS   TO 
IMPROPER    FRACTIONS 

ILLUSTRATIVE  EXAMPLE 
87.    Reduce  37f  to  an  improper  fraction. 

SOLUTION 

EXPLANATION.     8  eighths  =  1,  therefore, 
37  X  8  =  296  eighths     37  times  8  eighths,  or  296  eighths  must  equal 

296  +  5  =  301  eighths     37  times  1  or  37>    and  37*  must  e?ual  5 

eighths  more  than  296  eighths  or  301  eighths. 

Hence,  37|  =  ^ 


COMMON    FRACTIONS  53 

PRINCIPLES.  1.  One  integral  unit  is  expressed  as  a  fraction 
with  a  given  denominator  by  making  the  numerator  the  same  as  the 
given  denominator. 

2.  Two  or  more  integral  units  are  expressed  as  a  fraction  with  a 
given  denominator  by  making  the  numerator  as  many  times  the  given 
denominator  as  the  integer  is  times  1. 

RULE.  Multiply  the  integer  by  the  denominator  of  the  fraction, 
add  the  numerator  to  the  product,  and  the  result  will  be  the  numerator 
of  the  required  improper  fraction;  under  which  write  the  given 
denominator. 

NOTE.  When  no  particular  denominator  is  required,  an  integer  is  re- 
duced to  the  form  of  an  improper  fraction  by  simply  writing  1  underneath 
it  as  a  denominator  [Note,  77]. 

EXAMPLES   FOR  PRACTISE 
Reduce  the  following  to  improper  fractions : 

1.  9  to  16ths  5.   28*  9.   463B  13.   354flfc 

2.  75to8ths  6.   53J  10.   89H  14.   2186^ 

3.  61  7.   176A  11.   756J  16.   854* 

4.  12f  8.   253&  12.   817|  16.   612| 

REDUCTION    OF    IMPROPER    FRACTIONS    TO    INTEGRAL 
OR    MIXED    NUMBERS 

ILLUSTRATIVE  EXAMPLE 
88.   Reduce  -I3-  to  an  integer  or  mixed  number. 

EXPLANATION.     As  8  eighths  equal  one  integral  unit 
SOLUTION          [Prin.  1,  87],  there  must  be  as  many  integral  units  in 
173  _•_  g  =  21 5     173  eighths  as  8  eighths  are  contained  times  in   173 
eighths;    that  is,  21  integral  units  and  5  eighths. 

NOTE.  The  fractional  part  of  the  obtained  mixed  number  should  be 
reduced  to  lowest  terms,  if  not  already  so. 

EXAMPLES   FOR  PRACTISE 
Reduce  to  integers  or  to  mixed  numbers: 

1.  V-  4.  W         7.  W-  10.  MF          13. 

2.  H-8-          5.  W          8.  -HF          11.  *V*          14. 

3.  V-  6.  W  9.  *¥*  12.  Vff-          15. 


54  COMMON    FRACTIONS 

COMMON  DENOMINATOR 

89.  A  common  denominator  of  two  or  more  fractions  is  any 
one    denominator    to    which   each   of    them    can  be   separately 
reduced. 

90.  The  least  common  denominator  of  two  or  more  fractions 
is  the  least  denominator  to  which  each  of  them  can  be  separately 
reduced. 

NOTE.  A  fraction  can  be  reduced  to  any  higher  denominator  which  is 
a  dividend  of  its  own  denominator  [Note  1,  82].  A  common  denominator  of 
two  or  more  fractions  must  therefore  be  any  common  dividend  [60]  of  their 
several  denominators;  and  their  least  common  denominator  must  be  the  least 
common  dividend  [61]  of  their  several  denominators. 


REDUCTION  OF  FRACTIONS  TO  THEIR  LEAST  COMMON 
DENOMINATOR 

ILLUSTRATIVE    EXAMPLE 
91.   Reduce  f ,  f,  and  &  to  their  least  common  denominator. 

SOLUTION  EXPLANATION.     First    find    the    least 

L.  C.  D.  of  6,  8,  12  =  24     common  dividend  of  the  several  denomina- 
tors, as  shown  in  111.  Ex.,  63,  obtaining  24 
$    =  (24  -r-     6)  X  5  =  f  £     as  the  least  common  denominator. 

Then  reduce  each  fraction  to  the  ob- 
=  (^  ~     o)  X  o  =  sr     tained  least  common  denominator   (24)  as 

A  =  (24  +  12)  X  7  =  «    sho™  *  IU'  Ex"  82' 

MENTAL  PROCESS.  The  required  least  common  denominator  cannot 
obviously  be  less  than  the  greatest  of  the  given  denominators  (12);  and  it 
must  also  contain  all  the  factors  of  the  remaining  denominators  6  and  8 
which  are  not  found  in  12.  That  is,  12  contains  the  denominator  6  in  | 
and  it  contains  the  factor  4  of  the  denominator  8  in  f.  Hence,  incorporate 
with  12  the  remaining  factor  2  of  the  denominator  8  in  |,  and  12  X  2,  or 
24  is  the  required  least  common  denominator. 

RULE.  1.  Find  the  least  common  dividend  of  the  denominators 
of  the  given  fractions  as  shown  in  III.  Ex.,  63,  and  this  will  be  the 
required  least  common  denominator. 

2.  Reduce  each  of  the  given  fractions  to  this  obtained  least  com- 
mon denominator  as  shown  in  III.  Ex.,  82. 


COMMON    FRACTIONS  55 

NOTE  1.  If  the  denominators  of  the  several  fractions  have  no  common 
factor,  they  are  said  to  be  prime  to  each  other.  The  continued  product 
of  all  the  denominators  will  then  be  their  least  common  denominator  [Note 
1,  63]. 

NOTE  2.  If  any  given  denominator  is  seen  to  be  an  exact  divisor  of 
another  given  denominator,  it  need  not  be  considered  in  finding  the  least 
common  denominator  [Note  3,  63]. 

NOTE  3.  Before  attempting  to  find  their  least  common  denominator, 
the  several  fractions  should  be  reduced  to  lowest  terms,  if  not  already  so. 

EXAMPLES  FOR  PRACTISE 

Reduce  to  equivalent  fractions  of  the  least  common  denom- 
inator: 

1.  *,  !  9.  A,  H,  A 

2.  i  i  |  10.  7,  »,  f 

3.  A,  «,  *  H.  it,  4,  A 
4-  f,  I,  «  12.  |,  f ,  * 

6.  i,  !,  A  13.  I,  f ,  TV,  U 

6.  !,  i  f  U.  i  f,  f ,  H 

7.  * ,  A,  H  15.  i  f ,  f ,  f 

8.  f ,  A,  A  16.  i  9,  5,  H 


ADDITION    OF    FRACTIONS 

92.  Addition  of  fractions  is  the  process  of  finding  the  sum  of 
two  or  more  unequal  fractions. 

PRINCIPLES.  1.  Only  like  numbers  can  be  added  [Prin.  1,  18]. 
Therefore  the  several  numbers  of  fractional  units  (the  several  numera- 
tors, 72) ,  must  express  like  fractional  parts  (similar  denominators) . 

2.  Numerators  express  the  number  of  fractional  units  [72]  and 
denominators  simply  denote  the  names  of   those  fractional  units 
[73].     The  sum  of  two  or  more  fractions  must  therefore  be  the  sum  of 
their  respective  numerators. 

3.  The  sum  expresses  a  quantity  of  the  same  name  as  the  num- 
bers added  [Prin.  2,  18].      The  sum  of  the  several  numerators  must 
therefore  express  similar  fractional  units   (must  have  the  same 
denominator)  as  the  several  fractions  added. 


56 


COMMON   FRACTIONS 


TO  FIND  THE  SUM  OF  TWO  OR  MORE  FRACTIONS 

ILLUSTRATIVE    EXAMPLES 
1.   Find  the  sum  of  J,  f  ,  and  TV 

SOLUTION 

T    n    r>      *Q   Q    10       oxi 
.b.  ^.  JJ.  01  cJ,  8,  LZ  =  24 

4  -   (24    •    T)  V  1  8  f94tV^ 

8  (24ths) 

I  =  (24  -=-  8)  X  3  =     9  (24ths) 


=  (24  -i-  12)  X  7  =  14  (24ths) 
31 

=  1  2T> 


EXPLANATION.     First    reduce 
the  fractions  to  a  common  denom- 

mat°r  M  shown  m  I1L  Ex">  91' 
Then  add  the  resulting  numera_ 

t?rs  (8  +  9  +  14  =  3D,  and  under 
the  obtained  sum  (31)  place  the 
common  denominator  (24).  As 
the  result  (ii)  is  an  improper 
fraction,  reduce  it  to  a  mixed 
number  as  shown  in  111.  x.  88. 


2.   Find  the  sum  of  25J, 

SOLUTION 
L.  C.  D.  of  4,  12,  16  =  48 

251    =  (48  ^    4)  X     3  =  36  (48ths) 
91  fi  <L        (AQ    .    i<r»  \/     K        on  fAo±\.  \ 

216A  =   (48  -  12)  X    5  =  20  (48ths) 
73H  =  (48 


Ans. 


and  73H- 

EXPLANATION.  First  add 
the  fractions  as  if  there  were 
no  integers  (as  shown  in  pre- 
ceding  111.  Ex.  1),  obtaining 
%*>  or  1*»>  Write  the  frac- 

tional  part  of  the  obtained 

mfed  number  (n)  under  the 
column    of    fractions;     and 

carry  the  integral  part  of  the 
obtained  mixed  number  (1) 
to  the  units'  column  of  the 
Then  add  the  integers  (1,  the  carried  figure,  +  73  +  216  +  25), 


6)  X  11  =  33  (48ths) 

~89  —  141 
?^          ?8~ 


integers. 

obtaining  315fi  as  the  complete  sum. 


EXAMPLES  FOR  PRACTISE 


Find  the  sum  of 


+  I 
+  H 


2. 

3.  f  +  i  +  t 

4-  !  +  *  +  J 

6.  i  +  f  +  A 
6-  !  +  |  +  T6* 

7.  A  +  tt  +  S  +  t 

8.  f  +  J  +  f  +  A 

9.  *  +  H  +  *  +  A 


10. 
11. 

12. 

13. 
14. 
15. 
16. 
17. 
18. 


|  +  }  +  A  +  A 
I  +  f  +  f  +  f 

i  +  |  +  A  +  il 

8J  +  16*  +  27f  +  46f 


73$  +  27J  +  62|  +  91  A 
49§  +  56|  +  16J  +  39A 
628A  +  9H  +  28|  +  63  f 
19}  +  17f  +  26|  + 


COMMON    FRACTIONS  57 

NOTE  1.  It  is  usual,  in  the  dry-goods  business,  to  express  all  fractions 
of  a  yard  in  fourths.  The  denominator,  being  thus  understood,  is  not  expressed; 
and  the  numerator  is  written,  like  an  exponent,  at  the  right  of  the  integer. 

19.  Add  471  yards,  512  yards,  391  yards,  483  yards,  452  yards, 

46  yards. 

20.  Add  383  yards,  462  yards,  51  yards,  421  yards,  382  yards, 

47  yards,  39s  yards. 

NOTE  2.  It  is  customary  in  the  grocery  business  to  express  all  fractions 
of  a  pound  in  sixteenths,  or  ounces,  and  to  omit  writing  the  denominator 
understood. 

21.  Add  129  Ibs.,  353  Ibs.,  26  Ibs.,  188  Ibs.,  2311  Ibs.,  411 
Ibs.,  2316  Ibs. 

22.  Add  2735  Ibs.,  28612  Ibs.,  5638  Ibs.,  7253  Ibs.,  58613  Ibs., 
7511  Ibs. 

NOTE  3.  In  the  grain  business  fractions  of  a  bushel  are  similarly  ex- 
pressed in  GOths  (pounds)  for  wheat,  in  56ths  for  corn,  in  32nds  for  oats,  etc. 

23.  Add  the  following  weights  in  wheat:    75837  bu.,  64562 
bu.,  69843  bu. 

24.  How  much  corn  in  68739  bu.,  58723  bu.,  93418  bu.,  89642 
bu.? 

25.  How  much   oats  in  218312  bu.,  96827  bu.,  147230  bu., 
568319  bu.? 

SUBTRACTION    OF    FRACTIONS 

93.  Subtraction  of  fractions  is  the  process  of  finding  the  differ- 
ence between  two  unequal  fractions. 

PRINCIPLES.  1.  The  difference  can  be  found  only  between  like 
numbers  [Prin.  1,  22].  Therefore,  the  number  of  fractional  units 
(the  numerator)  of  the  subtrahend  must  be  of  the  same  name  (must 
have  the  same  denominator)  as  the  number  of  fractional  units 
(the  numerator)  of  the  minuend. 

2.  Numerators  express  the  number  of  fractional  units  [72]  and 
denominators  merely  denote  the  names  of  those  fractional  units  [73]. 
The  difference  between  two  fractions  must  therefore  express  the  differ- 
ence between  their  respective  numerators. 


58  COMMON   FRACTIONS 

3.  The  difference  between  like  numbers  must  also  be  like  [Prin. 
2,  22].  Therefore,  the  difference  between  two  numbers  of  fractional 
units  (two  numerators)  must  express  fractional  units  of  the  same 
name  (the  same  denominator)  as  those  of  the  subtrahend  and  min- 
uend. 

TO   FIND  THE  DIFFERENCE   BETWEEN  TWO  FRACTIONS 

ILLUSTRATIVE  EXAMPLE 
94.   Subtract  if  from 


SOLUTION  EXPLANATION.     First 

reduce  the  given  mixed 
L.C.D.  of  18  and  12  =  36  number  (2^)  to  an  im- 

2*  =  H  (36  +  12)  X  31  =  93(36ths)  %%%  fS^J^m. 

H  (36  +  18)  X  13  =  26(36ths)  ^  refuce  *?*?**: 

—  v  '  lent  fractions  of  the  least 

?5  ==  **•>  Ans.  common  denominator  as 
shown  in  111.  Ex.,  91, 
obtaining  ft  and  f  f . 

Next  subtract  the  less  numerator  (26)  from  the  greater  numerator  (93), 
obtaining  67  as  the  difference  between  the  two  numerators,  under  which  write 
the  common  denominator  (36ths)  to  denote  the  name  of  this  difference. 
As  the  result  (ft)  is  an  improper  fraction,  reduce  to  a  mixed  number  as 
shown  in  111.  Ex.,  68,  obtaining  Ifi. 

NOTE.  If  the  mixed  number  contains  few  integral  units,  reduce  it  to  the 
form  of  an  improper  fraction  before  performing  the  subtraction;  but  if  it  con- 
tains many  integral  units,  apply  the  method  shown  in  95. 

EXAMPLES  FOR  PRACTISE 

Find  the  difference  between 

1.  |  and  it  9.  §  and  J 

2.  §  and  }  10.  H  and  M 

3.  f  and  H  11.  I  and  1 

4.  f  and  |  12.  J  and  \\ 

5.  J  and  1  13.  2\  and  3i 

6.  1  and  |  14.  If  and  2TV 

7.  T%  and  H  15.  2TV  and 

8.  f  and  f  16.  2|  and  3J 


COMMON    FRACTIONS  59 

SUBTRACTION   OF   MIXED   NUMBERS 

ILLUSTRATIVE  EXAMPLES 
95.  1.   Subtract  75  from  87J. 

SOLUTION  EXPLANATION.     First  find  the  fractional  part  of 

87-         3(4ths)      tne  remamder  by  subtracting  no  fourths  in  the  sub- 
__  P.  ,  ..,    ^     trahend  from  3  fourths  in  the  minuend,  obtaining  f. 

Next  find  the  integral  part  of  the  remainder  by 

12  f         3(4ths)     subtracting  the  integral  part  of  the  subtrahend  (75) 
from  the  integral  part  of  the  minuend  (87),  obtaining 
12,  thus  making  the  complete  remainder  12  f. 

2.  Subtract  216£  from  325. 

SOLUTION  EXPLANATION.     As   5   sixths   in   the   subtrahend 

4  .  cannot  be  subtracted  from  no  sixths  hi  the  minuend, 

325         6(6ths)      "borrow"  1  integral  unit  (  =  6  sixths)  from  the  integral 

91ft5       ^ffitV*  \      Part  °f  tne  minuend,  325.     Then  subtract  5  sixths  of 

"S)      the  subtrahend  from  the  borrowed  6  sixths  of  the 

108^       l(6th)       minuend,  obtaining  1  sixth  as  the  fractional  part  of 

the  remainder.     Next  subtract  the  integral  part  of  the 

subtrahend  (216)  from  the  integral  part  of  the  minuend  (324,  that  is,  325  less 
the  borrowed  1),  obtaining  108  as  the  integral  part  of  the  remainder,  thus 
making  the  complete  remainder  108&. 

3.  Subtract  176f  from  358|. 

SOLUTION  EXPLANATION.    First   reduce    the 

fractional  parts  of  the  subtrahend  and 

.L.U.L).  OI  4  and  o  =    \&  minuend   to  a  common  denominator 

<WM2  '  fiW  5  -  10(12tW>  <12ths)-  Then  subtract  9  twelfths  in 

the  subtrahend  from  10  twelfths  in 

176f  (12  -5-  4)  X  3  =  9(12th8)       the  minuend,  obtaining  1  twelfth  as 

182TVj  Ans.  l(12th)  the  fractional  part  of  the  remainder, 

to  which  prefix  the  difference  between 

the  integers  (182),  obtaining  182TV  as  the  complete  remainder. 

4.    Subtract  249H  from  368|.  EXPLANATION. 

SOLUTION  First     reduce     the 

fractions  to  a  com- 
L.C.D.  of  12  and  8  =  24  mon     denominator 


(24.    8)Xll  =  33(24ths) 
249H       (24  .  12)  X  11  =  ?2  (24ths)      the    minuend   and 

11  (24ths)       22     twenty-fourths 
in  the  subtrahend. 


60  COMMON    FRACTIONS 

As  22  twenty-fourths  in  the  subtrahend  cannot  be  taken  from  9  twenty- 
fourths  in  the  minuend,  borrow  1  integral  unit  from  the  integral  part  of  the 
minuend  (368).  Add  the  borrowed  integral  unit,  which  equals  24  twenty- 
fourths  to  the  9  twenty-fourths,  obtaining  367||  as  the  equivalent  minuend. 
Subtract  22  twenty-fourths  in  the  subtrahend  from  33  twenty-fourths  in  the 
minuend,  obtaining  1 1  twenty-fourths  as  the  fractional  part  of  the  remainder. 

Lastly,  subtract  the  integral  part  of  the  subtrahend  (249)  from  the  in- 
tegral part  of  the  minuend  (367,  that  is,  368  less  the  borrowed  unit)  obtaining 
118  as  the  integral  part  of  the  remainder,  thus  making  the  complete  remainder 
118H- 

EXAMPLES  FOR  PRACTISE 
Find  the  difference  between 

1.  617  and  928f  9.  52J  and  71f 

2.  248f  and  562  10.  341  and  63T% 

3.  325  and  7151  11.  28i|  and  57^ 

4.  183|  and  327  12.  65|  and  82  J 
6.  29f  and  86f  13.  24|  and  46| 

6.  42f  and  59J  14.   35^  and  64M 

7.  72i  and  92H  15.   48§  and  71T\ 

8.  23T3*  and  58f  16.   56i  and  84| 

MULTIPLICATION    OF   FRACTIONS 

96.  Multiplication  of  fractions  is  the  process  of  finding  the 
product  if  one  or  both  factors  are  fractions. 

NOTE.  If  the  multiplicand  denotes  the  value  of  one  (entire)  considered 
equal  quantity  [Note  1,  a,  25],  and  the  multiplier  denotes  the  number  of  such 
considered  quantities  [Note  1,  b,  25],  and  the  product  denotes  the  value  of 
all  such  considered  quantities  [Note  1,  c,  25],  it  must  follow  that  if  the  multi- 
plier is  1,  or  an  improper  fraction  equal  to  1,  the  product  must  exactly  equal 
the  multiplicand;  that  if  the  multiplier  is  a  mixed  number  or  an  improper 
fraction,  that  is,  if  the  multiplier  is  greater  than  1  and  therefore  denotes  that 
the  multiplicand  is  required  to  be  taken  more  than  one  time,  the  product 
must  be  greater  than  the  multiplicand;  and  if  the  multiplier  is  a  proper 
fraction,  or  less  than  1,  it  will  denote  that  the  multiplicand  is  required  to  be 
taken  less  than  one  time,  and,  therefore,  the  product  must  be  less  than  the 
multiplicand.  Hence, 

PRINCIPLE.  Multiplication  signifies  increase  when  the  multi- 
plier is  greater  than  I,  and  decrease  when  the  multiplier  is  less 
than  1. 


COMMON   FRACTIONS  61 

TO    MULTIPLY   ONE    FRACTION    BY    ANOTHER   FRACTION 
ILLUSTRATIVE  EXAMPLES 

97.  Multiply  (1)  !  by  |;  (2)  16  by  f ;  (3)  |  by  12;  (4)  2| 
by  A. 

SOLUTION  FIRST  OPERATION.     8  times  3  fourths  equal  24 

,*^  fourths;  therefore  one-ninth  of  8  times  3  fourths,  that 

is,  f  times  f ,  must  equal  $  of  ^,  or  ft  [Prin.  3,  26] 
£  V  I  =  II  =  -       wmch  in  lowest  terms  [84]  equal  f . 

SECOND  OPERATION.     As  the  numerators  3  and  8 
^r>  have  been  shown  in  the  first  operation  to  be  factors 

12  of  the  numerator  of  the  answer,  and  therefore  the 
^        ^        2  factors  of  the  dividend  of  an  unexecuted  division  [79], 
4        9  ~~  an(l  the  denominators  4  and  9  have  been  there  shown 

13  to  be  factors  of  the  denominator  of  the  answer,  and, 
therefore,  factors  of  the  divisor  of  the  same  unexecuted 

division  [79],  and  any  factor  common  to  both  divisor  and  dividend  may  be 
canceled  as  shown  in  111.  Ex.,  69,  it  will  be  found  more  convenient  first  to 
cancel  all  factors  common  to  opposite  terms  as  in  the  second  operation;  and 
then  multiply  the  uncanceled  factors  of  the  numerators  to  obtain  the  numer- 
ator of  the  product,  and  the  uncanceled  factors  of  the  denominators  to  find 
the  denominator  of  the  product. 

SOLUTIONS 

(2)  (3)  (4) 

2  4  1        4 

#  X  |  -  6  p  X  n  =  V  =  6f          |  X  I  =  | 

13  15 

The  explanation  of  Ex.  1  will  also  apply  to  Exs.  2  and  3,  as  will  become 
evident  by  conceiving  the  denominator  1  to  be  written  under  each  integral 
factor  [Note,  77]. 

RULE.  Multiply  the  numerators  of  the  fractions  to  obtain  the 
numerator  of  the  product,  and  multiply  their  denominators  to  obtain 
the  denominator  of  the  product,  first  canceling  all  factors  which  are 
common  to  opposite  terms. 

NOTE  1.  Before  multiplication,  mixed  numbers  which  express  few 
integral  units  may  be  reduced  to  improper  fractions  [87];  and  simple  integers 
should  be  regarded  as  numerators  with  the  denominator  1  understood  [Note, 
77]. 


62  COMMON   FRACTIONS 

NOTE  2.  If  one  of  the  factors  is  a  simple  integer  of  many  figures,  it  will 
usually  be  found  more  convenient  to  multiply  the  integer  by  the  numerator 
of  the  fraction,  and  to  divide  the  result  by  the  denominator. 

NOTE  3.  When  the  word  "of"  immediately  follows  a  fraction,  it  signifies 
times;  thus,  f  of  f  means  f  times  f ;  and  3  of  a  fortune  means  £  times  a  fortune, 
a  fortune  being  called  the  verbal  multiplicand  to  distinguish  it  from  the  ordi- 
nary numerical  multiplicand. 

EXAMPLES  FOR  PRACTISE 
Multiply  Multiply  Multiply 

1.  *  by  A  6.   &  by  5J  11.   673  by  } 

2.  f  by  |  7.   74  by  |  12.    f  by  276 

3.  f  by  A  8.   2f  by  }J  13.   783  by  2i 

4.  T3*  by  I  9.   2J  by  f  14.   Find  f  of  A 

5.  f  by  34  10.   6f  by  5i  15.   Find  f  of  f  of  T8* 

MULTIPLICATION   OF  INTEGRAL  AND   MIXED  NUMBERS 
ILLUSTRATIVE  EXAMPLES 

98.   Multiply  (1)  426|  by  8;   and  (2)  675  by  29}. 

SOLUTIONS  EXPLANATION.       (1)  8 

/j\  /2)  times    2    thirds    equal    16 

g7£  thirds,  or  5i     Write  $  as 

the  fractional  part  of  the 

426  f  ^9*         product,  and  carry  the  in- 

8  4)2025  tegral  part   (5)  to  the  in- 

o/i7oi  ~XM\ i         tegral  part  of  the  product 

OU°4          (426  X  8),  obtaining  3413 1 
b075  ^  t^  complete  product. 

(f  X  8  =  ^  =  5J)  1350  (2)  First  multiply  675 

2008 11         ky   the  fractional   part   of 
the  multiplier   (f),   as  di- 
rected in  Note  2,  97,  obtaining  506  j.     Then  multiply  675  by  the  integral 
part  of  the  multiplier  (29),  and  add  the  several  partial  products,  obtaining 
2008 lj  as  the  complete  product. 

EXAMPLES    FOR    PRACTISE 
Multiply  Multiply  Multiply 

1.  678|  by  7         5.   816  by  8}  9.   629  by  18* 

2.  295|by9         6.   295  by  6 f  10.   571  f  by  35 

3.  427f  by  8         7.   378  by  52J  11.   826f  by  42 

4.  572by3f         8.   483  by  25 }  12.    382|  by  85 


COMMON    FRACTIONS  63 

MULTIPLICATION   OF  MIXED  NUMBERS  BY  MIXED 

NUMBERS 

ILLUSTRATIVE  EXAMPLE 
99.   Multiply  79f  by  27  f. 

SOLUTION  EXPLANATION.      The   total    product 

must  equal  the  product  of  each  part  of 

79  f  the  multiplicand  separately  multiplied  by 

27 1  each  part  of  the  multiplier  [Prin.  2,  26]. 

3       2  I  _      £/9,« +v   \      Therefore,  multiply  the  fractional  part  of 

ns'      the  multiplicand    (f)   by  the  fractional 

79  X  f  =  52f  =  16(24ths)  part  Of  the  multiplier  (f),  obtaining  £. 
|  X  27  =  10J  =  _3(24ths)  Next  multiply  the  integral  part  of  the 
79  X  7  =  553  S  =  I.1  multiplicand  (79)  by  the  fractional  part 
79  v/  2  =  158  °^  tne  multipl*er  (s)>  obtaining  52f. 

1  Now   that  att  of  the  multiplicand  has 

been  multiplied  by  the  fractional  part  of 


2196^,  Ans.  the   multiplier,   separately   multiply  the 

fractional    and    integral    parts   of    the 

multiplicand  (|  and  79)  by  the  integral  part  of  the  multiplier  (27),  and  add 
the  several  partial  products  obtaining  2196s1!  as  the  complete  product  of  all 
the  orders  of  the  multiplicand  multiplied  by  att  the  orders  of  the  multiplier. 

RULE.  1.  Multiply  the  fraction  of  the  multiplicand  by  the 
fraction  of  the  multiplier.  2.  Multiply  the  integer  of  the  multipli- 
cand by  the  fraction  of  the  multiplier.  3.  Multiply  the  fraction  of 
the  multiplicand  by  the  integer  of  the  multiplier.  4.  Multiply  the 
integer  of  the  multiplicand  by  the  integer  of  the  multiplier.  5.  Add 
the  several  partial  products. 

EXAMPLES  FOR  PRACTISE 

Multiply  Multiply  Multiply 

1.  56fby36|  6.  286^  by  28£  11.   5684 f  by  28 A 

2.  42|  by  64J  7.  672|  by  40£  12.   9275£  by  93| 

3.  631  by  75|  8.  548!  by  36i  13.   3246J  by  74| 

4.  37A  by  231  9.  715|  by  12f  14.   4183f  by  300f 
6.   51|  by  18J  10.  368f  by  74J  15.   958U  by  673| 

DIVISION    OF   FRACTIONS 

100.  Division  of  fractions  is  the  process  of  dividing  when 
the  divisor  or  dividend,  or  both  of  them,  are  fractions  or  mixed 
numbers. 


64  COMMON    FRACTIONS 

NOTE.  If  division  is  the  opposite  process  to  multiplication  [Note  1,  48], 
Principle,  96,  taken  reversely,  will  apply  to  division;  that  is, 

PRINCIPLES.  1.  //  the  divisor  is  greater  than  1,  the  quotient 
will  be  less  than  the  dividend.  2.  //  the  divisor  is  less  than  1,  the 
quotient  will  be  greater  than  the  dividend. 


DIVISION  OF  FRACTIONS  BY  INTEGERS 

ILLUSTRATIVE  EXAMPLES 
101.  Divide  (1)  f  by  4;  (2)  f  by  3;   (3)  |  by  6;   (4)  3}  by  21. 

SOLUTIONS  EXPLANATION.     (1)  8  ninths  divided  by  4  equal 

/•n  2  ninths  [Prin.  1,  81].     For,  if  f  expresses  the  quotient 

g  _._  4        2  of  an  unexecuted  division  [79]  of  which  the  numerator 

=  Q,  Ans.       (8)  is  the  dividend  and  the  denominator  (9)  is  the 

divisor,  then  to  divide  the  dividend  (8)  by  4  is  equiva- 

(%\  lent  to  dividing  the  quotient  (the  fraction)  by  4  [Prin. 

5  5  4'34>- 

o  v/  o  =  ^r,  Ans.  (2)  As  the  numerator  (5)  is  not  divisible  by  the 

divisor  (3),  as  in  the  preceding  operation,  use  Prin. 

2  2,  81,  and  multiply  the  denominator  (8)  by  the  divisor 

2  (3),  obtaining  ^  as  the  quotient.     For,  to  multiply 

0  y  $  =  rv7>  Ans.     the  divisor  of  an  unexecuted  division  by  3  is  equivalent 

to  dividing  the  quotient  (the  fraction)  by  3  [Prin.  6,  34]. 
,£.  (3)  As  the  numerator  (4)  is  not  divisible  by  the 

1  entire  divisor  (6)  but  is  divisible  by  one  of  its  factors 
J                   1  (2),  divide  the  numerator  (4)  by  the  factor  2,  as  in 

2  V  21  =  6'  Ans.     Hi.  Ex.  1,  and  multiply  the  denominator  (9)  by  the 

3  remaining  factor  3  (2  X  3  =  6),  as  in  111.  Ex.  2,  obtain- 

ing tfV  [Summary,  1,  37]. 

(4)  First  reduce  the  small  mixed  number  (3?)  to  an  improper  fraction  (1). 
As  the  numerator  of  |  is  not  divisible  by  the  entire  divisor  (21)  but  is 
divisible  by  one  of  its  factors  (7),  divide  the  numerator  (7)  by  the  factor  7  of 
the  divisor  (21  =  7  X  3),  obtaining  1  as  the  new  numerator  and  \  as  the 
quotient,  then,  in  succession,  [Summary  1,  37]  divide  ^  by  the  remaining 
factor  of  the  divisor  (3)  as  already  shown  in  the  second  111.  Ex.,  obtaining 
i  as  the  final  quotient. 

RULE.  1.  Divide  the  numerator  of  the  fractional  dividend  by 
the  divisor,  and  let  the  denominator  remain  unchanged.  Or, 

2.  Multiply  the  denominator  of  the  fractional  dividend  by  the 
divisor,  and  let  the  numerator  remain  unchanged.  Or, 


COMMON    FRACTIONS  65 

3.  Divide  the  numerator  of  the  fractional  dividend  by  the  great- 
est factor  of  the  divisor  which  will  exactly  divide  the  numerator, 
and  multiply  the  denominator  of  the  fractional  dividend  by  the 
remaining  factor  of  the  divisor. 

NOTE.  If  the  dividend  is  a  mixed  number  which  contains  few  integral 
units,  it  should  be  reduced  to  an  improper  fraction. 


EXAMPLES   FOR  PRACTISE 
Divide  Divide  Divide 

1.  f  by  3  6.  ft  by  6  11.  4§  by  7 

2.  |  by  6  7.  5i  by  14  12.  6f  by  5 

3.  iJbyS  8.  6§by8  13.  A  by  20 

4.  I  by  16  9.  2i  by  9  14.  J  by  28 

5.  |  by  20  10.  f  by  8  15.  8J  by  14 

DIVISIONS  OF  INTEGERS  BY  FRACTIONS 

102.  Division  by  any  number,  integral  or  fractional,  is  the 
reverse  of  multiplication  by  the  same  number  [Note  1,  48];  that 
is,  8  divided  by  2  is  the  opposite  process  to  8  multiplied  by  2;  for 
in  the  former  8  is  made  one-half  as  great,  and  in  the  latter,  8  is 
inversely  made  twice  as  great.  Therefore,  any  division  by  a 
fractional  divisor  must  be  the  inverse  process  of  multiplication  by 
that  same  fractional  quantity.  Now,  in  multiplying  6  by  f ,  it 
has  already  been  seen  that  6  is  multiplied  by  the  numerator  (2) 
and  divided  by  the  denominator  (3).  Hence,  when  performing 
the  reverse  operation  of  dividing  6  by  f ,  the  reverse  process  should 
be  followed  and  6  should  be  multiplied  by  the  denominator  (3) 
and  divided  by  the  numerator  (2) ;  that  is,  in  multiplying  6  by  f , 
the  statement  is  6  X  f  [111.  Ex.  2,  97];  but  in  dividing  6  by  f, 
the  statement  should  be  6  X  i,  or  the  former  process  with  the 
terms  of  the  divisor  reversed.  Hence, 

SUMMARY.  1.  Division  by  a  fractional  divisor  is  the  opposite 
process  to  multiplication  by  a  fractional  multiplier. 

2.  Division  by  any  fractional  divisor  is  the  equivalent  of  mul- 
tiplication by  that  same  fractional  divisor  with  its  terms  inverted. 


66  COMMON    FRACTIONS 

ILLUSTRATIVE  EXAMPLES 
Divide  (1)  8  by  £;   (2)  15  by  3  J. 

SOLUTIONS 

EXPLANATION.     (1)  If  the  dividend  (8)  be  divided 

g  -i-  -  —  by  the  numerator  of  the  divisor  (4),  that  is,  by  five 

5  times  the  true  divisor  (|),  the  resulting  quotient  (2) 

2  g  will  be  one-fifth  of  the  true  quotient    [Prin.  6,  34]. 
$  X  ~i  =  10                Therefore,  five  times  2  or  10,  must  be  the  true  quo- 

tient; and  5  times  \  of  8  or  f  of  8,  that  is,  8  multiplied 

by  f  with  its  terms  inverted   (8  X  I)  must  be  the 

/o\  proper  process  for  dividing  8  by  f. 

*  '  (2)  The   quotient   of   15   divided   by   25   eighths 

tc  _i_?§_  (3  1)  is  the  same  as  the  quotient  of  one-fifth  of  15,  or  3, 

8  divided  by  one-fifth  of  25  eighths,  or  f   [Prin.  5,  34]; 

3  n  and  3  -T-  f,  as  shown  in  the  first  solution,  is  equal  to 


5 

RULE.     Multiply  the  dividend  by  the  fractional  divisor  with  its 
terms  inverted. 

EXAMPLES  FOR  PRACTISE 
Divide  Divide  Divide 

1.  6  by  f  5.  15  by  §f  9.   15  by  2£ 

2.  15  by  f  6.  2  by  f  10.   2967  by  f 

3.  24byf  7.  19  by  f  11.   6183  by  3| 

4.  8byf  .            8.  28  by  I  12.  7458  by  8  i 

DIVISION  OF  FRACTIONS  BY  FRACTIONS 
ILLUSTRATIVE  EXAMPLE 

103.   Divide  f  by  f  • 

SOLUTION  EXPLANATION.     If  8  ninths  be  divided  by  2,  which 

g  _i_  2   _  is  three  times  the  true  divisor  (f),  the  quotient  (f)  will 

be  one-third  of  the  true  quotient   [Prin.  6,  34].     There- 

fore 3  times  f  divided  by  2,  that  is,  three  times  one-half 

§$_4_1     of  f  ,  or  £  X  f  must  be  the  true  quotient,  and  this  is 

^$~~3=      *     equivalent  to  multiplying  the  dividend  by  the  divisor 

3        i  with  its  terms  inverted. 


COMMON   FRACTIONS  67 

RULE.     Multiply  the  dividend  by  the  fractional  divisor  with  its 
terms  inverted. 

EXAMPLES  FOR  PRACTISE 
Divide  Divide  Divide 

1.  f  by  f  6.  3f  by  f  9.  f  by  f 

2.  J  by  f  6.  8J  by  f  10.  31  by  4f 

3.  *by»  7.  Aby2|  11.  11  by  21 

4.  2iby|  8.  A  by  4}  12.  5J  by  3J 


DIVISION  OF  MIXED  NUMBERS  BY  INTEGERS 

ILLUSTRATIVE  EXAMPLE 
104.   Divide  919f  by  27. 

EXPLANATION.     First  divide  the  integral  part  of 

27)9191  (34  A  the  dividend  (919)  °y  the  divisor  (27),  obtaining  34 

gj  as  the  integral  part  of  the  quotient. 

Then  bring  down  the  fractional  part  of  dividend 

(I)  to  the  remainder  (1),  obtaining  1|  or  V;    and 

108  divide  this  as  yet  undivided  part  of  the  dividend  (V) 

j!  _  jus  by  the  divisor  (27),  obtaining  75j  as  the  fractional  part 

of  the  quotient,  thus  making  the  complete  quotient 


1>       1         5 

L¥-  x  —  =  —  If  desired,  7\  may  be  reduced  to  any  desired  de- 

8        ff       72  nominator  as  shown  in  111.  Ex.,  86. 

RULE.     Separately  divide  the  integral  and  the  fractional  parts 
of  the  dividend  by  the  divisor.     Or,  proceed  as  shown  in  105. 


EXAMPLES   FOR  PRACTISE 
Divide  Divide  Divide 

1.  8375f  by  15       4.   62515  by  45  7.   3265*  by  372 

2.  5268f  by  36       5.   9286f\  by  84         8.   58293A  by  173 

3.  6739|  by  28      6.   2480JI  by  16         9.   6875|  by  540 


COMMON    FRACTIONS 

DIVISION  OF  INTEGERS  BY  MIXED  NUMBERS 
105.   Divide  896  by  27  j. 

SOLUTION  EXPLANATION.     To    multiply   both    divisor 

27-)  896  and    dividend  by  the  same  number  does  not 

o         g  affect  the  value  of   the  quotient   [Prin.  5,  34]. 

Therefore,  multiply  both  divisor  (27|)  and  divi- 

221)7168(32^6T  dend  (896)  by  the  denominator  of  the  fractional 

663  part  of  the  divisor  (8),  obtaining  221  eighths 

roo  as  the  equivalent  divisor  and  7168  eighths  as 

the     equivalent     dividend.      Divide     as     with 

integers  [36].      If  necessary,  $&•  can  be  reduced 

96  to    an    approximate    fraction    of    any    desired 

96  —  221  =    9  6          denominator  by  86. 


RULE.  Multiply  both  divisor  and  dividend  by  the  denominator 
of  the  terminal  fraction,  and  proceed  with  the  results  as  in  integral 
division. 

EXAMPLES  FOR  PRACTISE 
Divide  Divide  Divide 

1.  6783by41f       4.   8729  by  328f         7.   28973  by  269* 

2.  5286by37|       5.   7123  by  416$         8.   35286  by  713t 

3.  2857  by  25f       6.   9285  by  876f         9.   41627  by  697J 

DIVISION  OF  MIXED  NUMBERS  BY  MIXED  NUMBERS 

ILLUSTRATIVE  EXAMPLE 
106.   Divide  684f  by  47J. 

SOLUTION 

EXPLANATION.     Multiply  both  divisor  and  divi- 
dend by  the  least  common  denominator  of  their 

285)4108(14H£  terminal  fractions  (L.  C.  D.  of  2  and  3  =  6),  obtain- 

2g5  ing  285  sixths  as  the  equivalent  divisor,  and  4108 

sixths   as   the   equivalent   dividend    [Prin.    5,    34]. 
1258  Divide  as  with  integers  [36]. 

1140  If  required,  ilf  can  be  reduced  to  an  approxi- 

mate  fraction  of  any  desired  denominator  by  86. 


118  -5-  285 


RELATION    OF    COMMON    FRACTIONS  69 

EXAMPLES   FOR  PRACTISE 


Divide 

Divide 

Divide 

1. 

695| 

by 

27| 

4. 

6395|  by  34} 

7. 

3486f 

by 

537J 

2. 

438^ 

by 

18* 

5. 

5284A  by  65J 

8. 

27581- 

by 

238f 

3. 

741  f 

by 

53f 

6. 

9172|  by  24i 

9. 

8175| 

by 

196^ 

Approximating  the  fractional  part  of  the  quotient  to  the  specified  de- 
nominator, divide 

10.  475f  by  73*,  to  16ths        13.   7283*  by  196f ,  to  4ths 

11.  9183|  by  29J,  to  8ths          14.    5976J  by  39J,  to  64ths 

12.  6198|  by  16f ,  to  32nds       15.   2743  J  by  216£,  to  halves 

RELATION  OF  COMMON  FRACTIONS 

107.  INTRODUCTION.  The  principles  which  govern  the  rela- 
tion of  integers  [48  to  53],  inclusive,  are  unlimited  in  their  applica- 
tion to  all  numerical  expressions,  whether  integral  or  fractional. 
It  only  remains  to  explain  the  fractional  form  which  the  multi- 
plier now  assumes,  and  to  harmonize  its  definition  [Note,  96] 
with  that  of  an  integral  multiplier  [Note  1,  b,  25].  A  number  is 
the  expression  of  a  quantity  by  the  use  of  figures  [3],  and  may  mean 
"how  much"  as  well  as  "how  many."  Hence,  f  of  a  pound  is 
as  much  a  number  of  pounds  as  1  pound  is,  or  as  2  or  more  pounds 
are.  The  only  difference  between  them  is  in  the  form  of  expres- 
sion; |  (3  fourths)  expressing  a  fractional  number  of  pounds, 
and  the  others  expressing  integral  numbers  of  pounds.  Hence, 

PRINCIPLES.  1.  The  multiplicand  denotes  the  value  of  one 
considered  equal  quantity,  whether  integral  or  fractional.  2.  The 
multiplier  denotes  the  integral  or  fractional  number  of  considered 
equal  quantities.  3.  The  product  denotes  the  value  of  all  the  inte- 
gral or  fractional  number  of  considered  equal  quantities. 

NOTE.  The  student  is  cautioned  against  confounding  the  expression 
"fractional  number  of  units,"  with  "number  of  fractional  units."  In  the 
numerical  expression  f ,  f  is  the  fractional  number  of  units,  and  the  numerator 
3  is  the  number  of  fractional  units,  or  fourths. 

ILLUSTRATIVE  EXAMPLES 

1.  Find  the  cost  of  f  of  a  yard  of  silk  at  80  cents  per  yard. 

2.  If  |  of  a  yard  of  silk  cost  60  cents,  what  is  the  price  per 
yard? 


70  RELATION    OF    COMMON    FRACTIONS 

3.  How  much  silk  at  80  cents  per  yard  can  be  bought  for  60 
cents? 

SOLUTION 

Factors  of  f  80  ff  =  cost  of  one  entire  yard  =  multiplicand  1  Factors  of 
dividend    1      f  =  fractional  number  of  yards  =  multiplier  J    product 
Dividend,  if  _         _  ^  rf  fract.onal  number  _  product  if 

given  required 

EXPLANATION.  In  Ex.  1,  as  80?f  is  the  price  of  one  (entire)  yard,  it  is  a 
multiplicand;  as  f  expresses  the  fractional  number  of  yards,  it  is  a  multiplier. 
Hence,  multiply  these  two  factors,  obtaining  60  j£  as  the  cost  of  all  the  frac- 
tional number  of  yards,  or  product. 

In  Exs.  2  and  3,  the  cost  of  all  the  fractional  number  of  yards,  or  product 
(60  ff),  is  given.  Hence,  in  Ex.  2,  divide  the  given  product  (60 £)  by  its  given 
multiplier  factor  (the  fractional  number  of  yards,  f )  to  obtain  its  required 
multiplicand  factor  (the  price  of  one  entire  yard,  80  jf);  and  in  Ex.  3,  divide 
the  given  product  (the  cost  of  all  the  fractional  number  of  yards,  60^)  by  its 
given  multiplicand  factor  (the  price  of  one  entire  yard,  80^)  to  obtain  its 
required  multiplier  factor  (the  fractional  number  of  yards,  f ). 

4.  If  a  man  can  walk  3|  miles  in  one  hour,  what  part  of  an 
hour  will  he  require  to  walk  2J  miles? 

EXPLANATION.  3£  miles  is  a  multiplicand  because  it  is  the  distance 
walked  in  one  entire  hour  [Prin.  1]  and  2\  miles  is  a  product  because  it  is 
the  distance  walked  in  all  the  required  fractional  number  of  hours  [Prin.  3]. 
Therefore,  divide  the  given  product  (!)  by  its  given  multiplicand  factor  (Jsa), 
to  obtain  its  required  multiplier  factor  (|  X  iV  =  f ),  or  the  fractional  number 
of  hours. 

EXAMPLES  FOR  PRACTISE 

108.  xl.  What  is  the  cost  of  f  of  a  gallon  of  syrup  bought  at 
$J  per  gallon? 

-*  2.   At  $f  per  yard,  how  many  yards  of  goods  can  be  bought 
for  $2i? 

3.   If  a  horse  is  fed  f  of  a  peck  of  corn  twice  every  day,  how 
much  corn  will  be  necessary  to  last  4J  days? 
>4/  How  much  butter  at  S|  per  pound  can  be  bought  for  $3? 

6.  If  the  speed  of  a  steamboat  is  15|  miles  per  hour,  in  what 
time  can  it  go  a  distance  of  9J  miles? 

6.  A  gardener  wishes  to  divide  2J  acres  of  land  into  16  equal 
flower  beds.  How  much  land  will  each  bed  contain? 


COMPARISON   OF   QUANTITIES  71 

7.  If  a  man  can  complete  f  of  a  task  in  -fy  of  an  hour,  in  what 
time  can  he  perform  the  entire  task?     [Explanation  of  Ex.  17,  54]. 

8.  If  8  men  can  complete  a  piece  of  work  in  f  of  a  working  day, 
how  many  working  days  will  be  required  for  one  man  to  com- 
plete it? 

9.  If  28  men  can  do  a  certain  piece  of  work  in  3  J  working  days, 
how  many  working  days  will  be  necessary  for  7  men  to  do  the 
same  piece  of  work? 

10.  If  there  are  146  days  in  f  of  a  year,  how  many  days  are 
in  ||  of  a  year? 

11.  The  distance  between  two  villages  is  18f  miles.     If  a  trav- 
eler has  walked  f  of  the  distance,  how  many  miles  has  he  yet  to 
walk? 

-f  12.  What  is  the  total  cost  of  J  of  a  pound  of  cheese  at  $J  per 
pound,  A  of  a  pound  of  butter  at  $f  per  pound,  and  9f  gallons 
of  syrup  at  $|  per  gallon? 

-^  13.  If  it  cost  $T3ff  to  buy  f  of  a  yard  of  goods,  how  many  yards 
Of  the  same  goods  can  be  bought  for  $6?? 

14.  An  estate  was  bequeathed  to  a  mother  and  her  five  chil- 
dren, the  mother  to  receive  one-third,  and  the  remainder  to  be 
equally  divided  among  the  children.     What  part  of  the  estate 
did  each  child  receive? 

15.  From    a    barrel    of    cider   containing  49 f   gallons,   there 
were  drawn  7|  gallons  at  one  time,  16|  gallons  at  another  time, 
5 1  gallons  at  a  third  drawing,  and  12  J  gallons  at  a  fourth.     How 
many  gallons  remained  in  the  barrel? 

16.  2J  times  17f  yards  of  goods  are  how  much  less  than  5? 
times  28  J  yards  of  the  same  goods? 

COMPARISON  OF  QUANTITIES 

109.  INTRODUCTION,  (a)  In  a  comparison  of  quantities 
whether  integral  or  fractional,  any  one  of  them  may  be  arbitrar- 
ily assumed  as  the  standard,  as  the  author  of  the  problem  may 
direct.  It  is  therefore  necessary  that  the  problem  should  indi- 
cate which  of  the  considered  quantities  is  to  be  accepted  as  the 
standard  of  comparison.  The  conventional  form  of  language  for 
arbitrarily  indicating  the  standard  is  the  expression  "of"  [Note  3, 
97]  placed  immediately  after  the  fractional  multiplier  and  immedi- 
ately before  the  standard  of  comparison,  or  multiplicand.  Thus, 


72  COMPARISON    OF    QUANTITIES 

in  the  completed  comparison,  $2400  are  f  of  $3600,  "$3600"  is 
the  indicated  standard  of  comparison  or  multiplicand  (because 
it  is  made  to  follow  "§  of"),  and  expresses  the  value  of  one  con- 
sidered quantity  [Prin.  1,  107];  §  is  the  indicated  multiplier  (be- 
cause it  immediately  precedes  "of  $3600"),  and  expresses  the  frac- 
tional number  of  times  the  considered  standard  which  is  expressed 
by  the  quantity  compared  with  the  standard  [Prin.  2,  107] ;  and 
$2400  is  a  product  because  it  is  the  quantity  which  is  placed  in 
comparison  with  the  standard  by  denoting  the  number  of  times 
(f )  the  standard  ($3600)  which  is  expressed  by  $2400,  the  quan- 
tity compared  with  the  standard.  That  is, "  $2400  are  f  of  $3600," 
simply  means  $240*0  are  f  times  $3600. 

(6)  If,  however,  the  standard  of  comparison  is  to  be  indicated 
with  the  view  of  increasing  it,  the  phrase  "more  than"  should 
immediately  follow  the  fractional  multiplier  and  precede  the  as- 
sumed standard  of  comparison.  Thus,  in  the  completed  compari- 
son "$840  are  f  more  than  $600,"  $600  is  the  indicated  standard 
of  comparison,  or  multiplicand  (because  it  is  thus  pointed  out  by 
being  made  to  follow  "  f  more  than  ") ,  and  expresses  the  value  of  one 
considered  equal  quantity;  f  is  the  indicated  multiplier  (because 
it  immediately  precedes  "more  than  $600"),  and  expresses  the 
fractional  number  of  times  the  standard  which  the  quantity  com- 
pared with  the  standard  is  more  than  that  standard;  and  $840 
is  the  quantity  compared  with  the  standard,  or  a  product,  because 
it  expresses  the  value  ($840)  of  all  the  considered  fractional  num- 
ber of  times  the  standard  (f  more  than  once  or  £  of  the  standard, 
that  is,  I  of  $600).  Expressed  in  more  familiar  terms,  "$840  are 
f  more  than  $600"  means,  "$840  are  f  of  (times)  $600,  added  to 
$600." 

(c)  The  standard  of  comparison  may  be  indicated  in  a  way 
to  show  that  it  is  to  be  decreased,  in  which  case  the  indicator 
"less  than"  should  follow  immediately  after  the  fractional  mul- 
tiplier and  immediately  precede  the  indicated  standard  of  com- 
parison, or  multiplicand.  Thus,  in  the  completed  comparison 
"$750  are  f  less  than  $1200,  the  standard  of  comparison,  or  multi- 
plicand, is  $1200;  f  less  than  once  (f)  the  standard,  or  f  times 
the  standard,  is  the  multiplier;  and  $750  is  the  quantity  compared 
with  the  standard  or  product.  In  other  words,  the  problem  means, 
$750  are  f  of  (times)  $1200,  subtracted  from  $1200. 

PRINCIPLES.  1.  The  multiplicand  is  the  value  of  one  indicated 
quantity  which  is  assumed  as  a  standard  with  which  some  other 
quantity  is  to  be  compared.  2.  The  multiplier  is  the  fractional  part 
of  the  standard  (the  fractional  number  of  times  the  standard) 


COMPARISON    OF    QUANTITIES  73 

which  is  expressed  by  the  quantity  compared  with  the  standard. 
3.  The  product  is  the  quantity  compared  with  the  standard  (the 
value  of  all  the  fractional  number  of  times  the  standard  which 
is  expressed  by  the  multiplier). 

NOTE  1.  A  problem  in  comparison  is  one  in  which  two  of  the  three 
terms  of  a  perfected  comparison  are  given,  and  it  is  required  to  find  the  omitted 
third  term. 

NOTE  2.  In  problems  in  which  the  numerical  standard  of  comparison 
is  the  omitted  third  term,  the  verbal  standard  will  have  to  be  substituted  after 
the  indicators  "of,"  "more  than,"  or  "less  than."  Thus,  in  the  problem: 
"If  200  acres  are  f  of  A's  farm,  how  many  acres  does  A's  farm  contain,"  "A's 
farm"  is  the  indicated  verbal  standard  of  comparison,  or  verbal  multiplicand 
which  is  required  to  be  expressed  numerically;  200  acres  is  the  quantity 
compared  with  the  standard  (A's  farm),  or  the  product;  and  f  is  the  multi- 
plier or  fractional  number  of  times  A's  farm  expressed  by  the  product  (200 
acres). 


ILLUSTRATIVE  EXAMPLES 

1.  What  part  of  $200  are  $80? 

SOLUTION  EXPLANATION.     $200  is  the  indicated  stand- 

ard of  comparison,  or   multiplicand;   $80  is  the 
=  Stfff  =  5      quantity  compared  with  the  indicated  standard,  or 
product.     Hence,  divide  the  given  product  ($80) 

by  its  given  multiplicand  factor  ($200),  expressing  the  unexecuted  division 
as  directed  in  Introduction,  a,  70,  obtaining  \  as  the  required  multiplier  fac- 
tor, or  fractional  number  of  times  (fractional  part  of)  the  standard  ($200) 
which  is  expressed  by  $80. 

2.  f  of  how  much  wheat  equal  \  of  a  bushel? 

SOLUTION  EXPLANATION.     "How  much  wheat"  is  the 

indicated  verbal  standard,  or  multiplicand,  of 
which  it  is  required  to  find  the  equivalent  numer- 

1  _._  4  =_}/L.—_  un      ical  expression  [Note  2,  above],     f  is  the  mul- 

^48  tiplier,  because  it  denotes  the  fractional  number 

2  of  times  the  standard  (how  much  wheat)  which 

is    expressed    by  the    given    product    (£    of    a 

bushel).  Hence,  divide  the  given  product  (i  bu.)  by  its  given  multiplier 
factor  (f),  obtaining  f  bu.  as  the  required  multiplicand  factor,  or  standard 
I  of  which  equal  i  bu. 


74  COMPARISON    OF    QUANTITIES 

3.   280  men  are  J  more  than  how  many  men? 

SOLUTION  EXPLANATION.       "How      many 

g  4  men"  is  the  indicated  verbal  stand- 

4  more  than  ?  =  ar(j  Of  comparison,  or  multiplicand, 

.„  of  which  it  is  required  to  find  the 

7  4  equivalent      numerical      expression. 

280  -f-  4  =  WP  X|  =  160  men        280  men  is  a   product,    because    it 

is  the  quantity  compared  with  the 
required    standard.       f    more    than 

once  the  required  standard,  that  is,  f  more  than  I  of  the  standard,  or 
|  of  the  standard,  is  the  multiplier,  because  it  denotes  the  fractional  num- 
ber of  times  the  standard,  or  multiplicand,  which  is  expressed  by  the  prod- 
uct (280  men)  or  quantity  compared  with  the  standard.  Hence,  divide  the 
given  product  (280  men)  by  its  indirectly  given  multiplier  factor  (1),  obtain- 
ing 160  men  as  the  required  multiplicand  factor,  or  standard  f  of  which  (£ 
times  which)  equal  280  men. 

EXAMPLES  FOR  PRACTISE 

110.M.   I  paid  $7200  for  a  farm  and  afterwards  sold  it  for 
|  of  what  I  paid  for  it.     How  much  did  I  receive  for  the  farm? 

2.  If  |  of  a  bushel  of  wheat  weigh  40  pounds,  what  is  the 
weight  of  an  entire  bushel? 

3.  What  part  of  $320  are  $180? 

4.  A  dealer  sold  f  of  his  stock  of  merchandise  for  $4824.     At 
how  much  was  the  entire  stock  valued? 

5.  A  man  bequeathed  &  of  an  estate  worth  $12352  to  a  daugh- 
ter.    How  much  did  she  receive? 

6.  48  bushels  of  corn  were  sold  from  a  bin  containing  56 
bushels.     What  part  of  the  contents  of  the  bin  was  sold? 

7.  In  a  certain  school  were  enrolled  246  boys  and  318  girls. 
What  part  of  the  total  enrolment  were  the  girls? 

8.  The  capital  of  a  firm  of  two  partners  was  $5600,  of  which 
one  partner  invested  $2100.     What  part  of  the  capital  did  the 
other  partner  invest? 

9.  A  grazier  sold  $  of  a  flock  of  sheep  and  had  114  sheep  re- 
maining.    How  many  sheep  did  he  sell? 

10.   A  man's  salary  is  $90  per  month  and  his  expenses  $60 
per  month.     What  part  of  his  salary  does  he  save? 
>     11.   A  is  worth  $13000,  or  f  more  than  B.     How  much  is  B 
worth? 


REVIEW   OF   FRACTIONS  75 

12.   A  man  paid  £  of  his  indebtedness  and  then  found  that 
he  still  owed  $7310.     How  much  did  he  originally  owe? 

REVIEW  OF  FRACTIONS 

111.     1.   What  is  the  sum  of  137f,  463f,  285J,  348TV,  825H, 
567  f,  and  362J? 

2.  A  man  engaged  to  plow  a  field  containing  78f  acres.     How 
many  acres  remained  unplowed  after  completing  f  of  his  engage- 
ment? 

3.  What  is  the  cost  of  58f  yards  of  goods  at  37 i  cents  per 
yard? 

4.  A  dealer  sold  16j  yards  from  a  piece  containing  38y5s  yards. 
What  part  of  the  piece  of  goods  did  he  sell? 

5.  The  distance  between  two  harbors  is  847H  miles.     After 
a  ship  has  sailed  J  of  this  distance,  how  many  miles  remain  to 
complete  the  voyage? 

6.  What  is  the  total  cost  of  6J  pounds  beef  at  18  J  cents  per 
pound,  3J  pounds  mutton  at  13  f  cents  per  pound,  and  f  of  a 
pound  of  lard  at  11 J  cents  per  pound? 

7.  A  farm  consists  of  6  fields  containing  respectively  68 J, 
57 1,  65TW,  42|f,  52 1,  and  63|f  acres,  a  vegetable  garden  con- 
taining H  of  an  acre,  an  orchard  of  If  acres,  and  a  lawn  of  2£ 
acres.     If  it  was  bought  at  the  average  rate  of  $36  per  acre,  how 
much  did  the  farm  cost? 

8.  If  a  ship  completes  A  of  its  voyage  on  the  first  day,  ^  on 
the  second,  \  on  the  third  day,  and  is  then  1300  miles  from  its 
destination,  what  is  the  total  distance  which  it  has  already  sailed? 

9.  If  a  railroad  train  moves  with  the  average  speed  of  37$ 
miles  per  hour,  in  how  many  hours  can  it  traverse  718J  miles? 

10.  What  is  the  cost  of  685  J  tons  coal  at  $5.75  per  ton? 

11.  /?  of  a  day  is  f  of  what  part  of  a  day? 

12.  345  gallons  are  what  part  of  460  gallons? 

13.  A  grammar  and  speller  together  cost  92  cents,  and  the 
speller  cost  \  as  much  as  the  grammar.     What  was  the  cost  of 
each? 

14.  A  can  complete  a  job  of  work  in  8  days,  and  B  in  10  days. 
In  what  time  can  they  complete  the  task  by  working  together? 


76  REVIEW   OF   FRACTIONS 

EXPLANATION.  If  A  can  complete  the  entire  job  in  8  days,  he  can  per- 
form I  of  the  job  in  1  day;  and  if  B  can  complete  the  entire  job  in  10  days, 
he  can  perform  rff  of  the  job  in  1  day.  Therefore,  by  working  together,  A 
and  B  can  perform  i  of  the  job  plus  yV  of  the  job,  or  &  of  the  job  in  1  day. 
That  is,  &  of  the  job  is  the  multiplicand,  because  it  expresses  the  value  of  one 
equal  day's  labor  performed  by  both  men  [Prin.  1,  107];  f  o  of  the  job  is  the 
product  understood,  because  it  expresses  the  value  of  all  the  equal  days'  labor 
performed  by  both  men.  Hence,  divide  the  product  understood  (?{})  by  its 
given  multiplicand  factor  (?%),  obtaining  4|  days  as  the  required  multiplier 
factor,  or  number  of  times  one  day's  labor  of  both  men  (&  of  the  job)  which 
will  equal  the  total  labor  of  both  men  (?$  of  the  job),  or  the  entire  job. 

15.  A  can  do  a  piece  of  work  in  5  days  and  B  in  7  days.     In 
how  many  days  can  they  do  the  piece  of  work  by  uniting  their 
labor? 

16.  A  can  perform  a  certain  task  in  6  days,  B  in  9  days,  and 
C  in  12  days.     In  what  time  can  the  task  be  completed  if  they 
work  together? 

17.  A  and  B,  working  together,  can  finish  a  piece  of  work  in 
12  days.     If  A  working  alone  can  do  it  in  20  days,  in  what  time  can 
B  do  it  without  any  help? 

18.  The  greater  of  two  numbers  is  347/j  and  the  less  265£. 
What  is  the  difference  between  them? 

19.  The  greater  of  two  fractions  is  ff,  and  their  difference  iV 
What  is  the  less  fraction? 

20.  The  less  of  two  numbers  is  758  f,  and  the  difference  between 
them  is  125£.     What  is  the  greater  number? 

21.  The  total  of  six  weighings  is  936  f  pounds.     If  five  of  the 
weighings  are  respectively  168  J,  271 J,  82^,  125f,  and  76  J  pounds, 
what  is  the  sixth  weighing? 

22.  If  coffee  loses  yV  of  its  weight  in  roasting,  how  much  green 
coffee  will  be  required  to  produce  375  pounds  of  roasted  coffee? 

23.  A  water  tank  has  two  outlets  of  different  diameters.     If 
only  the  smaller  outlet  is  opened,  the  tank  will  be  emptied  in  15 
minutes;  and  if  both  are  opened,  it  will  be  emptied  in  6  minutes. 
In  what  time  will  the  tank  be  emptied  if  only  the  larger  outlet 
be  opened? 

24.  A  merchant  withdrew  if  of  his  deposit  from  his  bank  and 
afterwards  had  $510  remaining  on  deposit.     How  much  did  the 
merchant  withdraw? 


DECIMAL    FRACTIONS  77 

25.  If  a  barrel  containing  28  gallons  of  vinegar  is  only  f 
full,  what  part  of  the  barrel  will  be  filled  if  7  gallons  of  vinegar 
are  added? 

26.  The  capital  of  a  firm  was  $24210.     If  the  investment  of 
one  partner  was  f  of  the  investment  of  the  other  partner,  what 
did  each  partner  invest? 

27.  A  man  bought  a  house,  paid  3!  of  the  purchase  money 
in  cash  and  still  owed  $5200.     What  was  the  cost  of  the  house? 

28.  Multiply  275f  by  38f  and  divide  the  product  by  18|. 

29.  Subtract  the  sum  of  268},  425£,  and  362f  from  the  sum  of 
349f,  562J,  416A,  and  183}. 

30.  A  man  owed  $3500  and  afterwards  paid  $1800  at  one 
time  and  $300  at  another  time.     What  part  of  his  original  debt 
did  he  still  owe? 

31.  A  invested  }  of  the  capital  of  a  firm,  B  },  C  },  and  D  the 
remainder.     If  D's  investment  was  $2782,  what  was  A's,  B's, 
and  C's? 

32.  A  man  bought  a  house  for  $8500,  paid  }  of  the  purchase 
money  at  one  time,  and  f  of  the  remainder  at  another  time.    How 
much  remained  unpaid? 

DECIMAL  FRACTIONS 

112.  INTRODUCTION.  The  new  form  of  fraction  now  to  be  con- 
sidered differs  from  that  of  common  fractions,  already  studied,  in 
this  one  respect,  that  the  denominators  of  decimals  are  not  usu- 
ally expressed,  whereas  in  common  fractions,  by  reason  of  the 
varying  scale  employed,  the  denominators  require  expression, 
except  in  a  certain  class  of  commercial  fractions  [Notes  1,  2,  3,  92]. 
The  distinguishing  characteristic  of  decimal  fractions  is  the  uni- 
form scale  of  ten  employed  in  expressing  the  equal  parts  into  which 
the  unit  of  the  fraction  [80]  is  divided,  instead  of  the  arbitrary, 
inconstant  scale  by  which  the  unit  of  common  fractions  may  be 
divided  into  any  number  of  equal  parts.  Reading  and  writing 
decimals  are  simply  extensions  on  a  descending  scale  of  the  manner 
of  reading  and  writing  integers.  Taking  the  unit  as  the  initial 
order,  the  first  figure  on  the  left  of  it  is  the  tens'  order,  and  the  first 
figure  on  the  right  of  it  is  the  tenths'  order;  the  second  figure  on 
the  left  of  the  units'  order  is  the  hundreds'  order,  and  the  second 
figure  on  its  right  is  the  hundredths'  order,  etc. 

While  the  methods  already  shown  for  operating  upon  common 


78  DECIMAL    FRACTIONS 

fractions  are  also  applicable  to  decimal  fractions,  if  the  latter  are 
made  to  assume  the  form  of  common  fractions  by  expressing  their 
denominators,  the  much  more  simple  and  convenient  methods  for 
reduction  of  integers,  and  for  adding,  subtracting,  multiplying,  and 
dividing  integers,  can  be  extended  to  decimal  fractions  by  reason 
of  the  same  decimal  system  of  notation  which  is  employed  for 
expressing  both. 

113.  A  decimal  fraction  expresses  one  or  more  of  the  decimal 
parts  of  an  integral  unit. 

Thus,  any  fraction  the  denominator  of  which  is  10,  or  the  product  of  10 
taken  twice  as  a  factor  (10  X  10  =  100),  or  the  continued  product  of  10  taken 
any  number  of  times  as  a  factor  (10  X  10  X  10  =  1000),  is  called  a  decimal 
fraction,  or  simply  a  decimal. 

114.  A  decimal  point  is  a  dot  (.)  placed  at  the  left  of  a  decimal 
to  indicate  its  denominator,  and  if  a  mixed  decimal,  to  separate 
the  fractional  from  the  integral  orders. 

The  denominator  of  a  decimal  is  understood  to  be  10  taken  as  many  times 
as  a  factor  as  the  number  of  places  which  the  right-hand  order  of  its  numera- 
tor is  located  to  the  right  of  the  decimal  point.  Thus,  in  .5,  the  numerator  5 
is  located  one  place  to  the  right  of  the  decimal  point  and  therefore  the  denom- 
inator is  understood  to  be  the  factor  10  taken  once;  in  .17  the  right-hand  order 
of  the  numerator  (7)  is  located  two  places  to  the  right  of  the  decimal  point, 
and  the  denominator  is  understood  to  be  the  factor  10  taken  two  times; 
that  is,  10  X  10,  or  100;  etc. 

115.  The  decimal  order  or   decimal  value   of  any  figure  of 
the  numerator  is  its  distance  from  the  decimal  point. 

The  nearer  any  order  of  the  numerator  is  to  the  decimal  point,  the  greater 
its  value,  because  10  will  be  a  factor  of  its  denominator  fewer  times,  thus 
dividing  the  unit  of  the  fraction  into  fewer  equal  parts,  and  correspondingly 
increasing  the  value  of  those  parts;  and  the  more  distant  any  order  of  the 
numerator  is  from  the  decimal  point,  the  less  its  value,  because  10  will  be  a 
factor  of  its  denominator  a  greater  number  of  times,  thus  dividing  the  unit  of 
the  fraction  into  a  greater  number  of  equal  parts,  and  correspondingly  diminish- 
ing the  value  of  those  parts. 

The  names  of  the  several  decimal  orders  and  their  relative 
values,  are  shown  in  the  following. 


| 

09 

ousands 

1 

Etc.,  etc. 
BILLIONS 

1 

1 

Ten-million 

MILLIONS 

Hundred-th 

Ten-thousa 

THOUSANDS 

Hundreds 

1 

| 

t> 

7 

3 

8 

2 

4 

9 

1 

5 

1 

6 

4 

jt 

-n 

^ 

JS 

4 

^ 

• 

\  2 

.- 

£ 

^ 

S 

S 

$ 

""O 

CO 

S 

-2| 

DECIMAL  FRACTIONS  79 

TABLE  OF  DECIMAL  VALUES 


I    i        .   1 

i  ii]ii  1 

•a  a  i  s  Y  -s  §  f  i 

ScGoa2  i     a 

'S^g^^KS^W 
7    1       8639475 


Integral  Orders  Decimal  Orders 

EXPLANATION.  In  considering  the  above  TABLE,  observe  1st,  that  the 
decimal  point  separates  the  decimal  orders  from  the  integral  orders;  2nd, 
that,  commencing  at  the  decimal  point,  the  integral  orders  are  enumerated 
from  right  to  left,  and  the  decimal  orders  from  left  to  right;  3rd,  that  the 
names  of  the  several  decimal  orders  correspond  with  those  of  the  integral 
orders  which  are  equally  distant  from  the  units1  order. 

116.  A  simple  decimal  is  one  which  is  composed  entirely  of 
decimal  orders,  having  no  integer  at  its  left,  and  no  common 
fraction  at  its  right;  as,  .175. 

117.  A  complex  decimal  is  one  which  has  a  common  fraction 
at  the  right  of  its  lowest  decimal  order;  as,  .286 f. 

NOTE.  A  complex  decimal  is  said  to  be  terminate  when  the  common 
fraction  at  its  right  can  be  expanded  to  an  equivalent  simple  decimal  [116] 
and  to  be  interminate  when  it  cannot  be  thus  expanded. 

118.  A  mixed  decimal  is  one  which  contains  integral  orders 
at  the  left  of  its  decimal  point;  as,  28.356. 

119.  The  unit  of  a  decimal  is  the  whole  unit  of  which  the 
decimal  expresses  a  part;  and  a  decimal  unit  is  one  of  the  equal 
decimal  parts  into  which  the  unit  of  the  decimal  has  been  divided. 

Thus,  in  the  expression  .25  of  a  yard,  one  whole  yard  is  the  unit  of  the 
decimal,  and  one  one-hundredth  of  a  yard  is  the  decimal  unit. 

120.  PRINCIPLES.     1.    The  denominator  of  a  decimal  is  1  fol- 
lowed by  a  cipher  for  every  place  that  its  right-hand  order  is  distant 
from  the  decimal  point. 


80  DECIMAL    FRACTIONS 

Thus,  in  .0375  the  denominator  is  1  followed  by  four  O's  (10000),  because 
its  right-hand  order  (5)  is  four  places  to  the  right  of  the  decimal  point;  and 
the  decimal  is  read  375  ten-thousandths. 

2.  Adding  ciphers  to,  or  taking  them  from  the  right  of  a  decimal, 
makes  no  change  in  its  value. 

To  place  O's  at  the  right  of  a  decimal  numerator  is  equivalent  to  multiply- 
ing both  the  numerator  and  the  denominator  by  10  for  every  0  annexed  [30]; 
and  to  omit  O's  from  the  right  of  a  decimal  numerator  is  equivalent  to  dividing 
both  the  numerator  and  the  denominator  by  10  for  every  0  omitted  [38] ;  and 
therefore,  in  neither  case,  does  it  change  the  value  of  the  fraction  [Prin.  4,  81]. 

3.  Prefixing  ciphers  to  a  decimal  numerator  and  moving  the 
decimal  point  to  the  left  of  the  prefixed  ciphers,  is  equivalent  to  divid- 
ing the  decimal  by  10  for  each  cipher  prefixed. 

To  place  O's  on  the  left  of  a  decimal  numerator  does  not  affect  the  value 
of  the  numerator,  but  increases  the  denominator  tenfold  for  each  0  so  prefixed 
[Prin.  1],  and  therefore  is  equivalent  to  dividing  the  fraction  by  10  for  each 
prefixed  cipher  [Prin.  2,  81]. 

4.  Moving  its  decimal  point  one  or  more  places  to  the  right  of 
its  former  position,  multiplies  the  decimal  by  10  for  each  place  so 
removed. 

To  move  the  decimal  point  to  the  right  of  its  former  position  will  bring 
the  right-hand  order  of  the  decimal  correspondingly  nearer  to  the  point,  and 
thus  diminish  the  number  of  O's  in  the  denominator  [Prin.  1] ;  but  to  diminish 
the  O's  in  the  denominator  is  equivalent  to  dividing  the  denominator  by 
10  for  each  0  rejected,  and  therefore  of  correspondingly  multiplying  the 
decimal  [Prin.  3,  81]. 

5.  Moving  its  decimal  point  one  or  more  places  to  the  left  of  its 
former  position,  divides  the  decimal  by  10  for  each  place  so  removed. 

To  move  the  decimal  point  to  the  left  of  its  former  position  will 
throw  the  right-hand  order  of  the  decimal  farther  from  the  point,  and  thus 
increase  the  number  of  O's  in  the  denominator  [Prin.  1];  but  to  increase 
the  O's  in  the  denominator  is  equivalent  to  multiplying  the  denominator 
by  10  for  each  additional  0,  and  therefore  of  correspondingly  dividing  the 
decimal  [Prin.  2,  81]. 


DECIMAL    FRACTIONS  81 

READING    DECIMALS 

121.  Read  the  decimal  .0289. 

EXPLANATION.  First  read  the  decimal  as  if  it  were  an  integer  (two  hun- 
dred eighty-nine);  then  name  its  denominator  (1  followed  by  four  ciphers,  or 
10000,  Prin.  1,  120). 

Read  the  following  decimals: 

1.  .7  7.  .375  13.  .06345  19.  57.816372 

2.  .09  8.  .0375  14.  .00918  20.  219.0071832 

3.  .005  9.  .58  15.  .07016  21.  8.00031678 

4.  .35  10.  .291  16.  .00053  22.  2.006521873 

5.  .046  11.  .8273  17.  .21897  23.  15.000005281 

6.  .673  12.  .625  18.  .00725  24.  7.8123768542 

WRITING  DECIMALS 

122.  Express  two  thousand  one  hundred  nine  hundred  thou- 
sandths by  the  use  of  figures. 

EXPLANATION.  The  denominator  (100000)  contains  five  ciphers,  and  the 
numerator  (2109)  contains  only  four  of  the  five  places  necessary  for  its  right- 
hand  order  (9)  to  be  removed  from  the  decimal  point  to  express  such  a  denom- 
inator [Prin.  1,  120].  Therefore,  write  the  decimal  point  and  one  cipher  for 
the  one  lacking  decimal  place  (.0),  and  then  annex  the  numerator,  obtaining 
.02109,  in  which  the  right-hand  order  (9)  is  located  five  places  to  the  right  of 
the  point,  and  is  thus  made  to  express  a  denominator  of  1  followed  by  five  O's, 
or  100000. 

NOTE.  The  number  of  ciphers,  if  any,  to  follow  the  decimal  point  (or 
to  be  prefixed  to  the  numerator)  will  always  be  the  number  of  ciphers  in  the 
denominator  diminished  by  the  number  of  figures  necessary  to  express  the 
numerator. 

Express  the  following  as  numerical  decimals : 

1.  Six  tenths  9.  328  ten-thousandths 

2.  Three  hundredths  10.  75  hundred-thousandths 

3.  Twenty-seven  hundredths  11.  4768  millionths 

4.  Five  thousandths  12.  27  ten-millionths 

5.  Forty-two  thousandths  13.  92536  millionths 

6.  Nine  hundredths  14.  527  hundred-thousandths 

7.  Sixty-three  thousandths  15.  8  millionths 

8.  Eighteen  ten-thousandths  16.  263  ten-millionths 


82  DECIMAL   FRACTIONS 

17.  Nineteen,  and  twenty-three  thousandths 

18.  Forty-six,  and  two  hundred  fifty-three  ten-thousandths 

19.  Ninety-five,  and  seventy-nine  thousandths 

20.  Six,  and  one  thousand  three  hundred  four  millionths 

21.  Two,  and  eight  thousand  ninety-six  hundred  thousandths 

22.  Three  and  four  thousand  seven  millionths 

23.  Four  hundred  fifty-three,  and  twenty-eight  thousand  seven 

hundred  forty-six  hundred-millionths 

24.  Eight  thousand  three  hundred  sixty-two,  and  one  hundred 

twenty-one  thousand  three  hundred  seventeen  billionths 

25.  Nine,   and  eighty-seven    million    two    hundred    ninety-four 

thousand  eight  hundred  seventy-one  billionths 

REDUCTION  OF  COMMON  FRACTIONS  TO  DECIMALS 

ILLUSTRATIVE  EXAMPLE 
123.   Reduce  T?S  to  an  equivalent  decimal. 

SOLUTION  EXPLANATION.     The  value  of  any  fraction  is  the 

1  fV>7  OOfW  437^  quotient  from  dividing  its  numerator  by  its  denomina- 
10;/.UUI  tor  [7Qj  Hence>  jf  10000  timeg  the  numerator  7  (or 

70000)  be  divided  by  the  denominator  (16),  the  result- 

60  ing  quotient  (4375)  will  be  10000  times  the  true  quo- 

4g  tient  [Prin.  4,  34].      Therefore  cut  off  four  places  from 

the  right  of  the  quotient  (.4375)  which  is  equivalent 

to  dividing  it  by  10000  [38],    thus  obtaining  the  true 

112  quotient. 

on  If  no  numerator  can  be  exactly  divisible  by  its 

«Q  denominator  except  it  contain  all  the  factors  of  that 

denominator  [Prin.  1,  62],  and  the  denominator  (16) 

contains  the  factor  2  four  times  (2X2X2X2  =  1 6), 

it  is  necessary  for  the  numerator  (7)  to  contain  the  factor  2  four  times  in 
order  to  be  exactly  divisible  by  the  denominator  (16).  As  annexing  one 
cipher  to  the  numerator  is  equivalent  to  multiplying  it  by  10  [30],  and  as 
10  contains  the  factor  2  once  (2X5  =  10),  it  follows  that  the  factor  2  is 
introduced  once  for  every  cipher  annexed  to  the  numerator,  and  therefore 
to  annex  four  O's  to  the  numerator  will  introduce  the  factor  2  four  times, 
and  thus  make  it  exactly  divisible  by  the  denominator. 

RULE.  Annex  ciphers  to  the  numerator  and  divide  by  the  denom- 
inator. Place  a  decimal  point  as  many  places  from  the  right  of  the 
quotient  as  there  have  been  ciphers  annexed  to  the  numerator. 


DECIMAL    FRACTIONS  83 

NOTE  1.  The  number  of  ciphers  to  be  annexed  to  the  numerator  will  be 
equal  to  the  greatest  number  of  factors  2  or  5  in  the  denominator.  A  common 
fraction  in  lowest  terms  cannot  therefore  be  reduced  to  a  simple  decimal  [116] 
if  its  denominator  contains  other  prime  factors  than  2  or  5,  because  annexing 
O's  to  the  numerator  cannot  introduce  such  other  factors  without  which  exact 
division  of  the  numerator  by  the  denominator  is  impossible  [Prin.  1,  62]. 

NOTE  2.  Common  fractions  should  be  in  lowest  terms  before  they  are 
reduced  to  decimals. 

NOTE  3.  Because  they  recur  so  frequently,  learners  should  memorize 
the  decimal  equivalents  of  halves,  fourths,  and  eighths.  \  =  .5;  i  =  .25; 
|  =  .75;  |  =  .125;  f  =  .375;  f  =  .625;  J  =  .875. 

EXAMPLES  FOR  PRACTISE 
Reduce  the  following  to  decimals: 

1.  i        4.   i        7.   &        10.  ft        13.   ft          16.   18T3* 

2.  f        5.    1        8.   ft        11.   H        14.   AV        17.  465ft 

3.  f        6.   |        9.  I          12.   |          16.  «          18.  671H 

NOTE  4.  Inconvenient  long  divisions  may  frequently  be  avoided  by 
separating  the  common  fraction  into  two  factors,  the  decimal  equivalent  of 
one  of  which  is  already  known  [Note  3].  Thus,  if  =  |  of  V;  and  ^  =  2f 
or  2.375;  therefore  \-  of  V  must  equal  1  of  2.375. 

Reduce  the  following  to  decimals  by  factoring : 

19.  M  22.   H  25.   f  J  28.   63H 

20.  H  23.   ft  26.   ft  29.   37ft 

21.  {$  24.   «  27.   58A  30.   78ft 

NOTE  5.  It  is  usually  sufficient  for  the  requirements  of  business  to  extend 
a  decimal  result  to  a  certain  number  of  decimal  places,  ordinarily  not  more 
than  three  or  four;  that  is,  as  many  ciphers  are  annexed  to  the  numerator  as 
there  are  decimal  places  required  in  the  result;  and  when  the  last  quotient 
figure  has  been  obtained,  to  increase  it  by  1  if  the  final  remainder  is  at  least 
one-half  the  divisor. 

Reduce  the  following  to  decimals  of  the  number  of  places 
indicated : 


2  dec.  pi. 

3  dec.  pi. 

4  dec.  pi. 

5  dec.  pi. 

6  dec.  pi. 

31.  168A 

34.  911J 

37.  231* 

40.  8.26  fr 

43.  1.2764  A 

32.  34T** 

35.  26ft 

38.  71*1 

41.  7.348| 

44.  3.296? 

33.  72§| 

36.  49A 

39.  3715 

42.  9.527f 

45.  2.83451 

84  DECIMAL    FRACTIONS 

REDUCTION  OF  DECIMALS  TO  COMMON  FRACTIONS 

ILLUSTRATIVE  EXAMPLES 
124.   Reduce  .57^  and  .1875  to  common  fractions. 

SECOND    SOLUTION 
FIRST    SOLUTION  1875    -^-  625  =     3 

57|  X  7  =  400  =  4  '1875  =  15555  -  625  =  16 

100  X  7  =  700       7  Or, 

1875=  11X8  =  15-5  =  ^ 

10  X  8  =  80  -5-  5  =  16 

FIRST  EXPLANATION.  Change  the  complex  decimal  to  the  form  of  a 
complex  common  fraction  by  expressing  its  decimal  denominator  100  (f^). 
Then  multiply  both  terms  of  the  obtained  common  fraction  by  the  denom- 
inator of  the  complex  numerator  (7)  to  obtain  an  equivalent  simple  common 
fraction  f°$  [Prin.  4,  81].  Finally,  reduce  to  lowest  terms  [84],  obtaining  f. 

SECOND  EXPLANATION.  Change  the  decimal  to  the  form  of  a  common 
fraction  (rVirw).  Then  reduce  to  lowest  terms  [84]. 

Or,  applying  Note  3,  123,  it  is  seen  that  the  last  three  figures  of  the  given 
decimal  (875)  equal  £.  Hence,  .1875  equal  j|.  Then  proceed  as  in  the  first 
solution. 

NOTE.  Business  men  do  not  usually  take  the  unnecessary  trouble  to  find 
the  greatest  common  divisor  as  was  done  in  the  second  solution;  but  as  5  and 
2  are  the  only  prime  factors  of  a  decimal  denominator,  they  successively  divide 
by  5  until  that  factor  has  been  eliminated  and  then  similarly  by  2;  or,  to 
avoid  many  petty  divisions  by  2,  they  divide  by  4  or  by  8,  if  seen  to  be  com- 
mon to  both  terms. 

RULE.  Omit  the  decimal  point  and  prefixed  ciphers  from  the 
given  decimal ;  express  the  remaining  figures  in  the  form  of  a  com- 
mon fraction  by  writing  the  denominator  understood ;  and  reduce  to 
lowest  terms  as  in  84. 

EXAMPLES  FOR  PRACTISE 
Reduce  the  following  to  common  fractions: 

1.  .75  6.  .6875  11.  .59375  .  16.  2.00875 

2.  .375  7.  .9375  12.  .00625  17.  19.66| 

3.  .005  8.  .0025  13.  5.0375  18.  6.333J 

4.  .125  9.  .15625  14.  17.0075  19.  9.714f 

5.  .0875  10.  .34375  15.  8.28125  20.  16.272A 


DECIMAL    FRACTIONS  85 

APPROXIMATION  OF  DECIMALS  TO  COMMON  FRACTIONS 
ILLUSTRATIVE  EXAMPLE 

125.   Approximate  .75173  to  a  common  fraction  with  a  denom- 
inator not  greater  than  eighths. 

SOLUTION  EXPLANATION.     The  unit  of  the  given  decimal,  ex- 

75173  pressed  in  eighths  equals  8  eighths,  therefore  .75173  of 

the  unit  of  the  fraction  must  equal  .75173  of  (times) 
8    eighths,    or    6.01384   eighths.     Hence,    .75173  =  f, 


6.01384  =  f  =  f    approximately,  or  in  lowest  terms, 

RULE.  Multiply  the  given  decimal  by  the  required  denominator. 
Point  off  as  many  places  in  the  product  as  are  found  in  the  given 
decimal.  If  the  decimal  part  of  the  product  equals  one-half  or  more, 
increase  the  integral  part  of  the  product  by  1.  The  integral  part 
of  the  product  will  be  the  required  numerator,  under  which  write 
the  given  denominator. 

EXAMPLES  FOR  PRACTISE 
Approximate  to  common  fractions  of  the  required  denominator : 

1.  .6879  to  16ths     4.   .6247  to  16ths     7.   .5168  to  32nds 

2.  .4912  to  32nds     5.    .7987  to  5ths       8.    .6672  to  3ds 

3.  .7489  to  8ths       6.    .2531  to  4ths       9.    .8334  to  6ths 

ADDITION  OF  DECIMALS 
ILLUSTRATIVE   EXAMPLES 
126.   1.   Add  .375,  2.75,  .0058,  15.5  and  .0625. 

SOLUTION         EXPLANATION.     Only  like  orders  can  be  added    [Prin.   1, 
18].     Therefore  arrange  the  decimals  to  be  added  so  that  their 
.o/O  points  shall  fall   in   an  upright   column,   thus  causing  orders 

2.75  which  are  at  the  same  distance  from  the  decimal  point  and 

.0058         which  must  consequently  be  like    [115],  to    fall  in  the  same 
15.5  column.     Add  the  columns  in  the  usual  manner.     Place  the 

Ofi25        decimal  point  in  the  sum  directly  under  the  points  of  the  num- 
bers added,  for  the  sum  of  any  column  must  be  of  the  same 
18.6933         order  as  that  of  the  column  added  [Prin.  2,  18],  and  must  there- 
fore be  located  at  the  same  distance  from  the  decimal  point  as 
the  column  added,  to  be  properly  indicated. 


86  DECIMAL   FRACTIONS 

2.   Add  19.29f,  .5f,  3.06if,  and  25/5,  to  4  decimal  places. 

SOLUTION 

iqoQoyr  EXPLANATION.     First   expand   the  complex   decimals 

[117]  and  mixed  numbers  [78]  to  five  decimal  places  [Note 

.OC  )O7  -  5^  123].     The  expression  to  one  decimal  place  more  than 

3.06813  —  the  required  number  is  intended  to  insure  accuracy  in  the 

25.09375  four  places  required  in  the  answer  by  obtaining  the  neces- 

,10  nooon  sary  carrying  figure   (2)  from  the  fifth  decimal  column. 

"iO.U^ZOU  m,  jj  j  •     ,         rf  .,  ,. 

Ihen  arrange,  add,  and  point  off,  as  in  the  preceding 
48.0223,  Ans.    example. 

NOTE.  If  the  rejected  figure  at  the  right  of  any  answer  is  5  or  more,  the 
last  retained  figure  should  be  increased  by  1.  That  is,  if,  in  the  111.  Ex., 
48.02237  had  been  the  obtained  total,  48.0224  would  have  been  a  closer 
approximation  to  this  exact  result  than  48.0223. 

EXAMPLES  FOR  PRACTISE 
Find  the  sum  of 

1.  37.41,  5.8967,  .3562,  863.7,  and  16.25. 

2.  2.183,  .713,  .005,  43.81,  2.1875,  and  32.025. 

3.  682.5,  6.073,  .058,  25.305,  97.1834,  and  5.07168. 

4.  29.38,  1.35867,  26.038,  534.7,  6.28734,  and  59.0625. 
6.  .0635,  .0072,  .9,  .85,  3.15825,  .075,  and  .6. 

6.  32.05i,  7.08A,  563f|,  16.07f,  and  6.9 J,  as  a  simple  decimal. 

7.  .08U,  .72f,  .84i,  46TV,  28.3  J,  and  A,  as  a  simple  decimal. 

8.  .62J,  J,  42 J,  .07T3ir,  48¥V,  and  .25Tf y,  as  a  simple  decimal. 

9.  52.6H,  7.82A,  .5|,  .17f,  and  61.8-f,  to  3  decimal  places. 

10.  7t,  -5J,  .68A,  2.91  A-,  -781,  and  J},  to  4  decimal  places. 

11.  .516f,  6.8f ,  12.343TV,  .6f,  and  f ,  to  5  decimal  places. 

12.  Sixteen,   and  one   hundred  sixty-four  ten  thousandths; 
two   hundred  twenty-eight,   and  three  thousand  five    hundred 
seventy-two  millionths;    forty-seven,  and  twenty-one  ten-thou- 
sandths;   eight  hundred  thirty-seven,  and  fifteen  thousand  one 
hundred  ninety-three  millionths. 

13.  Four  hundred  nineteen,  and  six  thousand  one  hundred 
fifty-six  hundred-thousandths;   twenty-three,  and  eighteen  thou- 
sand nine  hundred  seventy-four  ten-millionths ;   eight,  and  forty- 
two  thousand  eight  hundred  fifty-one  millionths;    one  hundred 
thirty-five    thousand    six    hundred    eighty-seven,    and    sixteen 
hundred-thousandths. 


DECIMAL   FRACTIONS  87 

14.  Twenty-three  millionths;  seven,  and  one  thousand  four 
hundred  thirty-two  hundred-thousandths;  fifty-nine,  and  thirteen 
thousand  seven  hundred  twenty-eight  ten-millionths;  eight  thou- 
sand three  hundred  fifty-eight  ten-thousandths;  seventy-five, 
and  eighty-six  thousandths;  nine  thousand  two  hundred,  and 
seven  thousand  three  hundred  nineteen  millionths. 

SUBTRACTION  OF  DECIMALS 
ILLUSTRATIVE  EXAMPLE 

127.   Subtract  68.25352  from  183.6|,  to  3  decimal  places. 
SOLUTION  EXPLANATION.     First  expand  each  decimal  to  four 

1 83  fi8cn  -4-  decimal  places,  or  to  one  more  place  than  is  required,  as 

explained  in  111.  Ex.  2,  126;   and  so  arrange  them  that 

the  decimal  point  of  the  subtrahend  shall  fall  directly 
115.4317  underneath  that  of  the  minuend,  thus  causing  similar 

orders  to  fall  in  the  same  column  [Prin.  1,  22].  Subtract 
115.432  — ,  Ans.  in  the  usual  manner.  Place  the  decimal  point  in  the 

result  directly  under  those  of  the  minuend  and  sub- 
trahend, because  the  difference  between  like  orders  must  also  be  like 
[Prin.  2,  22],  and  must  therefore  be  located  at  the  same  distance  from 
the  decimal  point  [115]  to  indicate  this  likeness. 

EXAMPLES   FOR  PRACTISE 
Find  the  difference  between 

1.  75.106  and  28.32  6.  43.007  and  18.26 

2.  837.25  and  256.3845  7.  528.1586  and  271.03 

3.  72.68  and  34.2478  8.  68  and  .3129 

4.  39.658  and  4.8  9.  42.175  and  28 
6.  76.23  and  .847  10.  .75  and  .21867 

11.  53. 28 f  and  27.2 A,  expanded  to  simple  decimals 

12.  268.4J  and  125.92|,  expanded  to  simple  decimals 

13.  9.68§|  and  2.934|,  expanded  to  simple  decimals 

14.  67.2f  and  39.16|,  to  3  decimal  places 

15.  95.13f  and  27.9F\,  to  4  decimal  places 

16.  18J  and  12J,  to  2  decimal  places. 

17.  Subtract  three  hundred  twenty-eight,  and  fifty-six  ten- 
thousandths  from  five  hundred  sixty. 

18.  Subtract  two  thousand  five  hundred  seventy-three  hun- 
dred-thousandths from  eight-tenths. 


88  DECIMAL    FRACTIONS 

MULTIPLICATION  OF  DECIMALS 

ILLUSTRATIVE  EXAMPLE 
128.   Multiply  9.865  by  2.45. 

SOLUTION  EXPLANATION.     If  the  multiplicand   (9.865)   be  expressed 

g  ogc  in    the    form    of    an    equivalent    improper    common    fraction 

'  (ftM);    and   if    the    multiplier    (2.45)    be   similarly   expressed 

^•4o  (fof);    and  their  product  obtained  as  common  fractions  by 

49325  97,  the  numerator  of  the  product  would  be  the  product  of 

39460  these  numerators   (9865  X  245,   or  2416925),   or  the  product 

1Q730  °^  t^ie  severa>l  01>ders  of  the  decimal  multiplicand  and  decimal 
multiplier  (9.865  X  2.45  =  2416925);    and   the   denominator 


24.16925  of  the  product  would  be  the  product  of  these  denominators 
(1000  X  100  =  100000),  that  is,  1  followed  by  as  many 
ciphers  as  are  found  in  both  factors,  or  five  ciphers  [34] ;  and  as  each  cipher 
in  the  denominator  requires  a  decimal  place  in  the  numerator  to  express 
it  [Prin.  1,  120],  cut  off  five  decimal  places  from  the  right  of  the  product 
(24.16925)  to  indicate  that  the  denominator  is  1  followed  by  5  ciphers,  or 
100000. 

RULE.  So  place  one  factor  under  the  other  that  the  right-hand 
order  of  the  lower  factor  shall  fall  under  the  right-hand  order  of  the 
upper  factor.  Multiply  as  in  integral  multiplication.  Point  off 
from  the  right  of  the  product  as  many  decimal  places  as  are  found 
in  both  factors. 

NOTE  1.  In  arranging  the  factors  for  multiplication,  their  decimal  points 
need  not  fall  in  an  upright  column,  as  in  addition  and  subtraction  of  decimals. 

NOTE  2.  If  any  obtained  product  has  fewer  figures  than  are  required  to 
be  pointed  off,  prefix  ciphers  to  supply  the  deficiency. 

NOTE  3.  To  multiply  a  decimal  by  10,  100,  1000,  etc.,  move  its  decimal 
point  as  many  places  to  the  right  as  there  are  ciphers  in  the  multiplier,  annex- 
ing ciphers  to  the  decimal  if  necessary. 

NOTE  4.  Complex  decimals,  if  terminate  [Note,  117],  should  be  expanded 
to  simple  decimals  before  multiplication;  and  if  interminate,  should  be  mul- 
tiplied as  shown  in  98  or  99. 

EXAMPLES  FOR  PRACTISE 

Multiply  Multiply  Multiply 

1.  516.78  by  43         4.  2.086  by  .034        7.  .685  by  .0058 

2.  79.832  by  7.4        5.  56.8  by  .0096        8.  7.6f  by  5.2 

3.  56.283  by  .59        6.  8.27  by  1.038        9.  6.28 A  by  .76 


DECIMAL    FRACTIONS  89 

Multiply  Multiply  Multiply 

10.  75.1ft-  by  .2f    14.  2.41/^by  1.6U       18.  2.34J  by  .16? 

11.  6.8£by3.6        15.  2.89|  by  3.5  19.  .7183 by  100  [Note  3] 

12.  2.96|  by  5.12i  16.  .684|  by  2.63         20.  .62  by  1000 

13.  6.73fby.913£  17.  .715J  by  3.8|  [99]  21.  .7183  by  10 

DIVISION   OF  DECIMALS 
ILLUSTRATIVE  EXAMPLE 
129.   Divide  10.02677  by  2.873. 

SOLUTION  EXPLANATION.     If  the  dividend  be  expressed 

o  07QMA  r»OA77^Q  AC\     m  ^e  ^orm  °*   an   e(luivalent   improper   common 

fraction  «WW),    and   the   divisor   be   similarly 

expressed    (|ff&$),    the    quotient    obtained    as    in 

1  4077  common  fractions  [103]  would  be  VoV&k7-  X  iff! , 

1  1492  or     218°7°3z6i7o7o;      that     is,     the     quotient    of    the 

dividend  regarded  as  an  integer  (1002677)  divided 

25857  by   the    divisor    regarded    as    an   integer    (2873), 

25857  obtaining  349,  and    this   result  (349)  divided    by 

100  [38],  or  by  1  followed  by  as  many  ciphers  as 

the  number  of  decimal   places  in  the  dividend  (five)  exceed  the  number 
of  decimal  places  in  the  divisor  (three),  obtaining  3.49. 

DEDUCTIONS.  1.  If  the  dividend  is  the  product  of  the  divisor  and  quo- 
tient (Note  2,  35),  then  the  dividend  (10.02677)  must  contain  as  many  deci- 
mal places  as  the  divisor  (2.873)  and  the  quotient  (3.49)  together  [Note, 
128].  Therefore,  if  the  divisor  contains  only  three  of  the  five  decimal  places 
found  in  the  dividend,  the  quotient  must  contain  the  remaining  two  decimal 
places.  Hence,  subtract  the  number  of  decimal  places  in  the  divisor  from  the 
number  of  decimal  places  in  the  dividend  to  find  the  number  of  decimal  places  to 
point  off  in  the  quotient. 

2.  If  the  dividend  contains  as  many  decimal  places  as  both  the  div'sor 
and  the  quotient,  it  must  contain  at  least  as  many  as  the  divisor.  Hence,  if, 
before  division,  any  dividend  is  seen  to  contain  fewer  decimal  places  than  its 
divisor,  annex  as  many  ciphers  to  such  a  dividend  as  will  supply  the  deficiency. 

RULE.  (1)  See  that  the  dividend  contains  at  least  as  many 
decimal  places  as  the  divisor.  (2)  Divide  as  with  whole  numbers. 
(3)  From  the  right  of  the  quotient  point  off  as  many  decimal  places 
as  the  number  of  decimal  places  in  the  dividend  exceeds  the  number 
of  decimal  places  in  the  divisor. 

NOTE  1.  If  the  quotient  does  not  contain  as  many  figures  as  it  must 
have  decimal  places,  prefix  ciphers  to  supply  the  deficiency. 

NOTE  2.     To  divide  a  decimal  by  10,  100,  1000,  etc.,  move  its  decimal 


90  DECIMAL    FRACTIONS 

point  as  many  places  to  the  left  of  its  former  position,  as  there  are  ciphers  in 
the  divisor,  prefixing  ciphers  to  the  dividend  if  necessary. 

NOTE  3.  Complex  decimals,  if  terminate  [Note  117],  should  be  expanded 
to  simple  decimals  before  commencing  the  division;  and  if  interminate,  should 
be  divided  as  shown  in  105  and  106. 

NOTE  4.  To  obtain  an  approximate  quotient  of  a  given  number  of  deci- 
mal places,  use  as  many  figures  after  the  decimal  point  of  the  dividend  as  will 
equal  the  number  of  decimal  places  in  the  divisor  plus  the  number  of  decimal 
places  required  in  the  quotient,  omitting  the  remaining  decimal  places  of  the 
dividend  if  it  has  an  excess,  or  annexing  decimal  ciphers  to  the  dividend  if  it 
has  a  deficiency,  of  decimal  places. 

EXAMPLES  FOR  PRACTISE 
Divide  Divide 

1.  16.341  by  3  9.  95  by  .0005 

2.  138.1716  by  49.347  10.  .0002512  by  .628 

3.  32.50676  by  56.83  11.  9.72  by  .00036 

4.  2175.4754  by  842  12.  .001  by  .000008 

5.  37  by  .0008  13.  1 53.8671  f  by  37.8  J  [Note  3] 

6.  .0366234  by  .537  14.  .5224  by  .006| 

7.  .00003  by  .005  15.  16.47H  by  3.7J 

8.  .000219936  by  68.73  16.  11.7196|  by  .016| 

Applying  Note  4,  divide 

17.  19.3  by  17  (to  4  dec.  places) 

18.  43.5782  by  3.4  (to  2  dec.  places) 

19.  .268345  by  5.3  (to  3  dec.  places) 

20.  5.93472  by  .26  (to  2  dec.  places) 

21.  617.35  by  48.3  (to  3  dec.  places) 

22.  .18T5s-  by  .348  (to  3  dec.  places) 

23.  394.8f  by  .417  (to  2  dec.  places) 

24.  58.52f  by  .78  (to  3  dec.  places) 

25.  16.7|  by  .18f  (to  4  dec.  places) 

26.  7.5J  by  13|  (to  2  dec.  places) 

Applying  Note  2,  divide 

27.  283.45  by  100  31.  .6|  by  10  (to  3  dec.  pi.) 

28.  4167.8  by  10  32.  1.7|  by  100  (to  6  dec.  pi.) 

29.  52.384  by  1000  33.  38.74  by  400  (=  100  X  4) 

30.  3.4|  by  100  34.  763.4  by  7000  (=     1000  X  7) 


DECIMAL    FRACTIONS  91 

WHEN  THE  DIVISOR  IS  A  SIMPLE  DECIMAL  OF   MANY 
PLACES 

130.  It  is  frequently  necessary  in  commercial  calculations 
to  divide  by  a  simple  decimal  of  many  places.  This  will  require 
much  unnecessary  labor  if  the  methods  of  129  are  employed. 
The  following  contraction  is  absolutely  correct,  and  will  often 
save  more  than  half  the  labor  required  by  the  ordinary  processes. 

ILLUSTRATIVE  EXAMPLE 

Divide  $5.86  by  .621783  to  2  decimal  places  (cents). 

SOLUTION  EXPLANATION.     First  see  that  the  divi- 

9  -U    dend    contains    no    grater    or    less    number 
than    four    decimal    places,    cutting    out    all 
beyond  the  fourth  place  if  it  has  more,  or 
2640  annexing  ciphers  to  supply  the  deficiency  if  it 

2487 


Then  divide  by  as  many  figures  of  the 
divisor    (.621783)    as   this   dividend   of   four 
124  decimal  places  (5.8600)  will  contain,  that  is, 

29  by  .6217$$,  canceling  the  remaining  figures  of 

the  divisor  as  unnecessary  to  secure  the  cor- 

rect quotient.  Dividing  this  adapted  dividend,  it  is  found  that  58600 
ten-thousandths  contain  6217  ten-thousandths  9  (integral)  times  with  a 
remainder  of  2640  ten-thousandths. 

Then  dropping  one  figure  (7)  from  the  right  of  the  preceding  divisor 
(62  1/),  it  is  seen  that  the  remainder  from  the  preceding  division  (2640  ten- 
thousandths)  contains  621  thousandths,  4  (tenths)  times,  with  a  remainder 
of  153  ten-thousandths. 

Again  dropping  one  figure  (1)  from  the  right  of  the  preceding  divisor 
(.62J),  it  is  seen  that  the  remainder  from  the  preceding  division  (153  ten- 
thousandths)  contains  62  hundredths,  2  (hundredths)  times.  As  the  quo- 
tient is  required  to  two  decimal  places,  or  to  hundredths,  the  division  is  now 
completed. 

RULE.  1.  First  see  that  the  dividend  contains  exactly  four 
decimal  places,  annexing  ciphers  if  it  has  a  deficiency,  or  rejecting 
all  beyond  the  fourth  place  if  it  has  an  excess. 

2.  At  the  first  partial  division,  use  as  many  of  the  higher  orders 
of  the  divisor  as  the  modified  dividend  of  four  decimal  places  can 
contain,  and  reject  the  remaining  orders. 


92  REVIEW   OF    DECIMALS 

3.  At  each  successive  partial  division,  reject  one  additional  order 
from  the  right  of  the  immediately  preceding  divisor.     The  last  partial 
division  should  be  by  the  number  expressed  by  the  first  two  figures 
after  the  decimal  point  of  the  divisor. 

4.  //,  at  any  stage  of  the  division,  the  remainder  does  not  contain 
the  divisor,  write  a  cipher  as  the  next  quotient  figure,  reject  one  more 
order  from  the  right  of  the  divisor,  and  proceed  as  before. 

NOTE  1.  When  multiplying  by  the  successive  quotient  figures,  always 
include  one  or  two  figures  next  adjoining  the  proper  divisor,  to  secure  the 
correct  carrying  figure  to  the  expressed  product. 

NOTE  2.  If  the  final  remainder  is  one-half  or  more  of  the  final  divisor, 
increase  the  final  quotient  figure  by  1. 

NOTE  3.  The  above  method  of  division  will  also  be  found  useful  when  a 
divisor  is  the  integer  1  followed  by  many  decimal  places.  In  this  case,  see 
that  the  dividend  contains  exactly  three  decimal  places,  instead  of  the  four 
places  required  when  the  divisor  is  a  simple  decimal.  The  final  divisor 
should  be  the  two  highest  orders  as  before,  but  these  two  orders  (regarded  as 
one  quantity)  will  now  express  tenths. 

EXAMPLES   FOR  PRACTISE 
Carrying  the  quotients  to  2  decimal  places,  divide 

1.  $9.16  by  .48736291  6.  $3.167894  by  .7264583 

2.  $12.50  by  .9376284  7.  $7.25  by  1.3427862  [Note  3] 

3.  $35.72  by  .8675913  8.  $45.6  by  1.2783245 

4.  $138.65  by  .9163748  9.  $25  by  1.6824357 

6.  $400  by  .28683297  10.  $18.67  by  1.2354987 

REVIEW  OF   DECIMALS 

131.  The  relation  of  mixed  decimals  [118]  to  each  other 
the  same  as  that  of  integers  [48-53];  and  the  relation  of  simple 
decimal  quantities,  the  same  as  that  of  proper  fractions  [107- 
109]. 

1.  What  is  the  difference  between  .087  times  72.5634  and 
the  sum  of  8.73,  16.035,  12.4875,  and  15.3? 

2.  The  average  yield  per  acre  of  a  field  of  wheat  was  32.1875 
bushels.     What  was  the  total  yield  if  the  field  contained  75.0625 
acres? 

3.  What  is  the  sum  of  15.6fV,  38.24^,  39|,  265,  f ,  and  2.34J, 
to  four  decimal  places? 


REVIEW    OF    DECIMALS  93 

4.  How  many  32nds  are  in  .34378?  [125] 

5.  What  is  the  cost  of  37  barrels  of  vinegar,  averaging  39.0625 
gallons  per  barrel  at  16 f  cents  per  gallon? 

6.  The  product  of  two  decimals  is  11392.5625  and  one  of 
its  factors  is  156.0625.     What  is  its  other  factor? 

7.  What  decimal  part  of  a  yard  of  cloth  can  be  bought  for 
$.60,  at  $.80  per  yard? 

8.  How  much  corn  in  .75  of  850  bushels? 

9.  A  and  B  own  a  tract  of  land,  A  owning  .625  of  the  tract. 
What  decimal  part  of  the  tract  does  B  own? 

10.  A  merchant  sold  .375  of  his  stock  of  flour.     How  many 
barrels  of  flour  did  he  originally  have  in  stock  if  the  quantity 
sold  was  162  barrels? 

11.  After  paying  .1875  of  his  debts,  a  merchant  found  that 
he  still  owes  $7800.     What  was  his  original  indebtedness? 

12.  A  man  had  650  yards  of  silk,  and  at  various  times  sold 
481  yards.     What  decimal  part  of  the  silk  did  he  sell? 

13.  A  merchant  had  a  certain  sum  deposited  in  a  bank.     If 
he  withdrew  .35  of  his  deposit  to  buy  140  barrels  of  flour  at  $6 
per  barrel,  how  much  has  he  still  remaining  in  the  bank? 

14.  A  man  bequeathed  $28560  to  his  wife,  son,  and  daughter, 
leaving  .475  of  this  sum  to  his  wife,  .315  of  it  to  his  daughter,  and 
the  remainder  to  his  son.     How  much  did  each  receive? 

15.  A  man  bought  a  house  and  lot  for  $3500.     If  the  lot  cost 
.75  of  what  he  paid  for  the  house,  how  much  did  he  pay  for  each? 

16.  A  man  lost  .465  of  his  wealth  in  speculation,  and  then 
found  that  he  was  still  worth  $8774.     How  much  was  he  worth 
before  the  speculation? 

17.  $119.48  is  .45  more  than  how  much  money?  * 

18.  $28.90  is  .15  less  than  how  much  money? 

19.  A  man  has  $678.60,  which  is  .45  more  than  his  neighbor 
has.     How  much  has  his  neighbor? 

20.  Since  the  census  of  1900,  a  town  has  lost  .18  of  its  popu- 
lation.    If  the  census  of  1910  showed  a  population  of  4428,  how 
many  inhabitants  did  it  have  at  the  census  of  1900? 

21.  A  and  B  engage  in  partnership,  A  investing  $31188  and 
B  investing  $36612.     What  decimal  part  of  the  capital  did  each 
invest? 


94  COMPOUND    NUMBERS 

22.  A  farmer  sold  450  bushels  of  wheat  to  one  man  at  97 
cents  per  bushel,  and  .36  less  than  that  quantity  to  another  man 
at  95  cents  per  bushel.     How  much  did  he  receive  from  both 
sales? 

23.  A  man's  income  is  $2500  per  annum  and  his  expenses 
$1800  per  annum.     What  decimal  part  of  his  income  does  he 
save? 

24.  A  man  paid  $23.75  for  a  suit  of  clothes,  $33.25  for  an 
overcoat,  $3.80  for  a  hat,  and  had  .36  of  his  money  remaining. 
How  much  money  did  he  have  at  first? 

25.  A  man's  expenses  are  .78  of  his  income.     How  much  does 
he  save,  if  his  expenses  are  $1404  per  annum? 

COMPOUND  NUMBERS 

132.  INTRODUCTION,  (a)  Quantities  have  now  been  consid- 
ered the  various  orders  of  which  have  been  expressed  upon  a 
uniform,  ascending  scale  of  10,  commencing  with  the  unit  as  the 
initial  order,  as  with  integers  [9-13  and  17-69].  It  has  also  been 
seen  that  this  same  convenient  arrangement  of  orders  can  be 
extended  conversely  to  fractional  parts  of  the  initial  unit  by  express- 
ing them  upon  a  uniform  descending  scale  of  10,  as  with  decimal 
fractions  [112-131]. 

(6)  In  contrast  with  the  subdivisions  of  the  initial  unit  into 
convenient  decimal  fractional  parts,  has  also  been  placed  the 
more  inconvenient  divisions  of  the  integral  unit  upon  an  arbi- 
trary changeable  scale,  as  with  common  fractions  [70-111],  in 
the  consideration  of  which  increased  difficulty  was  experienced 
in  completing  the  relation  of  quantities  thus  expressed,  by  reason 
of  the  varying  scale  in  which  the  subdivisions  (denominators)  of 
the  mutually  related  terms  were  expressed. 
*  (c)  A  compound  number,  now  to  be  considered,  is  the  expres- 
sion of  an  integral  quantity  upon  the  same  arbitrary  varying  scale, 
as  that  of  common  fractions.  So  that,  as  common  fractions  are 
when  contrasted  with  decimal  fractions,  so  are  compound  inte- 
gral quantities  when  contrasted  with  simple  integral  quantities; 
the  scale  in  passing  from  one  denominate  order  to  the  next  higher 
or  lower  order  being  as  irregular  as  in  passing  from  a  common 
fraction  of  one  denominator  to  another  fraction  of  a  higher  or  lower 
denominator,  thus  requiring  the  application  of  the  same  prin- 
ciples of  reduction  in  the  one  as  in  the  other. 

(d)  The  law  of  increase  and  decrease  is  universal  in  its  appli- 


COMPOUND    NUMBERS  95 

cation  to  all  quantities,  in  whatever  form  they  may  be  expressed. 
The  processes  for  completing  the  relation  of  quantities  are  there- 
fore the  same  for  compound  integers  as  for  simple  integers  [48-53], 
and  the  same  for  common  fractions  [107-109]  as  for  decimals, 
though  the  method  for  executing  these  processes  may  seem  to  dif- 
fer with  the  form  in  which  the  related  terms  are  expressed.  Any 
seeming  difference  will  quickly  disappear  if  related  compound 
numbers  or  common  fractions  are  conceived  to  be  reduced  to 
equivalent  expressions  which  have  a  common  decimal  scale. 

133.  A  denominate  number  is  the  expression  of  a  quantity 
in  units  of  weight,  of  measure,  or  of  value. 

134.  A   simple    denominate  number   expresses  a   denominate 
quantity  in  units  of  a  single  denomination  or  name. 

Thus,  3  pounds  (units  of  weight),  5  bushels  (units  of  measure),  and  6 
dollars  (units  of  value)  are  simple  denominate  numbers. 

135.  A   compound   denominate  number,   or,   as    it   is   usually 
abbreviated,  a  compound   number,  is  the  expression  of   one  de- 
nominate  quantity  in  two  or  more  different  denominations  or 
subdivisional  names. 

Thus,  "5  gallons  3  quarts  1  pint"  is  a  compound  number,  because  it 
expresses  one  fluid  quantity  in  units  of  three  different  denominations. 


MEASURES   OF   VALUE 

136.  United  States  money  is  the  legal  currency  of  the  United 
States.     Scale:    10  cents   (f)  =  1  dime   (d.);    10  dimes  or  100 
cents  =  1  dollar  ($);   10  dollars  =  1  eagle  (E). 

For  convenience,  the  denomination  mill  is  frequently  used  to  express 
tenths  of  a  cent. 

137.  English  or  sterling  money  is  the  legal  currency  of  Great 
Britain.     Scale:  4  farthings  (far.)  =  1  penny  (d.) ;   12  pence  =  1 
shilling  (s);   20  shillings  =  1  pound  (£). 

138.  The  following  Table  exhibits  the  intrinsic  value  of  the 
several  units  of  currency  of  the  principal  nations  of  the  world, 
estimated  in  United  States  money,  as  proclaimed  by  the  Secretary 
of  the  United  States  Treasury  on  Jan.  1,  1912. 


96 


COMPOUND    NUMBERS 


Country 

Legal  Standard 

Monetary  Unit 

Value  in 
Terms  of  U.  S. 
Gold  Dollar 

Argentine  Republic 

Gold  

Peso  

$     965 

Austria-Hungary    .  . 

Gold  

Crown  . 

203 

Belgium  

Gold  and  silver  . 

Franc  

.193 

Brazil 

Gold 

Milreis 

546 

British  America   .  .  . 

Gold  

Dollar    . 

1  000 

Chili               

Gold  

Peso 

365 

Denmark  

Gold  

Crown   

268 

Egypt  
France                  .  . 

Gold  
Gold  and  silver  . 

Pound  (100  piasters)  . 
Franc 

4.943 
193 

German  Empire.  .  .  . 

Gold  

Mark  

.238 

Great  Britain    
India  (British)    .  . 

Gold  
Gold 

Pound  Sterling  
Rupee    .    . 

4.8665 
324  5 

Italy 

Gold  and  silver 

Lira 

193 

Japan       

Gold  

Yen    

498 

Mexico             .        .  . 

Gold      .    ...... 

Peso 

498 

Norway 

Gold 

Crown  . 

268 

Panama  

Gold  

Balboa  

1  000 

Peru                     .    .  . 

Gold  

Libra  .                  .... 

48665 

Philippine  Islands 

Gold 

Peso  . 

500 

Portugal    

Gold  

Milreis  

1.080 

-  Russia 

Gold      

Ruble  

.515 

Spain 

Gold  and  silver  . 

Peseta 

193 

Sweden 

Gold 

Crown 

268 

Switzerland  .          .  . 

Gold  

Franc  

193 

Turkey  . 

Gold  . 

Piaster  . 

.044 

MEASURES    OF   WEIGHT 

139.  Avoirdupois  weight  is   the   usual  weight  of   commerce. 
Scale:    16  ounces  (oz.)  =  1  pound  (lb.);    100  pounds  =  1  hun- 
dredweight (cwt.);  20  hundredweight  =  1  ton  (T.). 

140.  Troy  weight  is  used  by  jewelers  in  weighing  gold,  silver, 
and  other  precious  metals.     Scale:    24  grains  (gr.)  =  1  penny- 
weight (pwt.);  20  pennyweights  =  1  ounce  (oz.);   12  ounces  =  1 
pound  (lb.). 

141.  Apothecaries'  weight  is  used  by  apothecaries  and  physi- 
cians  in   weighing   the   non-liquid   ingredients   of   prescriptions. 
Scale:  20  grains  (gr.)  =  1  scruple  (3);   3  scruples  =  1  dram  (3); 
8  drams  =  1  ounce  (3);   12  ounces  =  1  pound  (lb.). 


COMPOUND    NUMBERS  97 

142.  Miscellaneous  table  of  weights:    2240  pounds  =  1   long 
ton;    100  pounds  of  grain  or  flour  =  1  cental;    100  pounds  of 
nails  =  1  keg;    196  pounds  of  flour  =  1  barrel;    200  pounds  of 
salted  beef,  pork,  or  fish  =  1  barrel. 

NOTE  1.     It  is  customary  to  allow  the  following  weights  to  1  bushel: 

Barley 48  Ib.  Ear  corn 70  Ib.  Potatoes 60  Ib. 

Beans 60  "  Corn  meal 50  "  Rye 56   " 

Buckwheat  ....  48  "  Cotton  seed  ...  30  "  Timothy  seed  ...  45   " 

Clover  seed. . .  .60  "  Oats 32   "  Wheat GO   " 

Shelled  corn  . .  .56  "  Peas 60  "  Wheat  bran  ...  .20   " 

NOTE  2.  Denominations  of  the  same  name  in  the  Troy  Scale  and  Apothe- 
caries' Scale  have  the  same  weight;  but  both  of  these  differ  in  weight  from 
denominations  of  the  same  name  in  the  Avoirdupois  Scale.  Thus,  there  are 
5760  Troy  grains  in  1  pound  Troy,  and  7000  Troy  grains  in  1  pound  Avoir- 
dupois; and  there  are  480  Troy  grains  in  a  Troy  ounce  (5760  gr.  -s-  12),  but 
only  4371  Troy  grains  in  1  ounce  Avoirdupois  (7000  gr.  -5-  16). 

MEASURES    OF    EXTENSION 

143.  Linear  measure  is  used   in  measuring  lengths,  widths, 
heights,  depths,  and  distances.     Scale:    12  inches  (in.)  =  1  foot 
(ft.);  3  feet  =  1  yard  (yd.);  5J  yards  =  1  rod  (rd.);  40  rods  =  1 
furlong  (fur.);   8  furlongs  or  320  rods  =  1  mile  (mi.). 

144.  Surveyors'  linear  measure  is  used  by  surveyors  in  measur- 
ing lengths  of  roads,  boundaries  of  land,  etc.     Scale:   7.92  inches 
(in.)  =  1   link   (1.);    100  links  =  1   chain  (ch.);    80  chains  =  1 
mile  (mi.). 

145.  Circular  measure  is  used  in  measuring  angles  and  arcs, 
(parts  of  a  circle) ;   and  by  navigators  in  estimating  latitude  and 
longitude.     Scale:    60  seconds  (")  =  1  minute  (');    60  minutes 
=  1  degree  (°)  ;  360  degrees  =  1  circle  (C),  or  circumference  of 
the  earth. 

MEASURES    OF    SURFACE 

146.  Square  measure  is  used   in  measuring  ordinary  areas, 
such  as  the  surfaces  of  boards,  of  painting,  plastering,  papering, 
paving,  etc.,  and,  unofficially,  of  land.     Scale:   144  square  inches 
(sq.  in.)  =  1  square  foot  (sq.  ft.);   9  square  feet  =  1  square  yard 


COMPOUND    NUMBERS 


(sq.  yd.);  30i  square  yards  =  1  square  rod  (sq.  rd.);   160  square 
rods  =  1  acre  (A);  640  acres  =  1  square  mile  (sq.  mi.). 

147.  Surveyors'    square    measure    is    used    by   surveyors    in 
measuring  land.     Scale:    10,000  square  links  (sq.  1.)  =  1  square 
chain  (sq.  ch.);    10  square  chains  =  1  acre  (A);    640  acres  =  1 
square  mile  (sq.  mi.). 

MEASURES    OF    VOLUME 

148.  Cubic  measure   is   used   in  measuring  the  contents  or 
volume  of  bodies.     Scale:    1728  cubic  inches  (cu.  in.)  =  1  cubic 
foot  (cu.  ft.);  27  cubic  feet  =  1  cubic  yard  (cu.  yd.). 

149.  Wood  measure  is  used  in  measuring  cord  wood.     Scale: 
16  cubic  feet  (cu.  ft.)  =  1  cord  foot  (cd.  ft.);   8  cord  feet  or  128 
cubic  feet  =  1  cord  (cd.). 

MEASURES    OF    CAPACITY 

150.  Liquid  measure  is  used  in  measuring  ordinary  liquids. 
Scale:    4  gills  (gi.)  =  1  pint  (pt.);    2  pints  =  1  quart  (qt.);    4 
quarts  =  1  gallon  (gal.). 

151.  Apothecaries'  fluid  measure  is  used  by  apothecaries  and 
physicians  in  measuring  liquid  medicines.     Scale:  60  minims  ("I) 
=  1  fluiddrachm  (f3);    8  fluiddrachms  =   1  fluidounce  (f  5);    16 
fluidounces  =  1  pint  (0). 

152.  Dry  measure  is  used   in  measuring  fruits,  vegetables, 
and  grains.     Scale:   2  pints  (pt.)  =  1  quart  (qt.);    8  quarts  =  1 
peck  (pk.)   4  pecks  =  1  bushel  (bu.). 

MEASURE    OF    TIME 

153.  Scale:  60  seconds  (sec.)  =  1  minute  (min.) ;   60  minutes 
=  1  hour  (hr.);   24  hours  =  1  day  (da.);   365  days  =  1  common 
year(yr.);  366  days  =  Heap  year;  100  years  =  1  century  (cent.). 

NOTE  1.     Leap  years  and  common  years  are  identified  as  follows: 

1.  Every  year  which  is  not  exactly  divisible  by  4,  is  a  common  year. 

2.  Every  year  which  is  exactly  divisible  by  4,  w  a  leap  year. 

3.  Centennial  years  (years  ending  in  two  ciphers)  when  divisible  by  400, 
are  leap  years;  and  when  not  divisible  by  400,  are  common  years. 


COMPOUND    NUMBERS  99 

NOTE  2.     The  order,  name,  and  length  in  days  of  each  month  are: 


Order  Name 

First January 

Second February 

Third March  . . 

Fourth April  . . . 

Fifth May 

Sixth   .       . .  June  . . 


Length 
. 31  days 
28  or  29  da, 
.31  days 
.30     " 
.31     " 
.30     " 


Order  Name  Length 

Seventh July 31  daya 

Eighth August 31      " 

Ninth September. . .  30     " 

Tenth October 31      " 

Eleventh November  ...  30     " 

Twelfth  ....  December  ...  31      " 


OTHER    MEASURES 

154.  In  enumerating  certain  articles  of  merchandise,  the  fol- 
lowing Scale  is  used:    12  things  =  1  dozen  (doz.);    12  dozen  =  1 
gross  (gro.);   12  gross  =  1  great  gross  (g.  gro.);  6  things  =  1  set; 
20  things  =  1  score. 

155.  In  measuring  paper,  the  following  Scale  is  used:    24 
sheets  (sh.)  =  1  quire  (qu.);  20  quires  =  1  ream  (rm.);   2  reams 
=  1  bundle  (bdl.);   5  bundles  =  1  bale  (bl.). 

REDUCTION  OF  COMPOUND  DENOMINATE  NUMBERS  TO 
SIMPLE  DENOMINATE  NUMBERS 

ILLUSTRATIVE  EXAMPLE 
156. 


SOLUTION 

15  da.  17  hr.  28  min. 
X  24    (hrs.  in  1  da.) 
377  (15  X  24  +  17) 
X60  (min.  in  1  hr.) 


Reduce  15  da.  17  hr.  28  min.  to  seconds. 

EXPLANATION.  Commence  with  the 
highest  denomination  (15  da.)  and  reduce  it 
to  the  next  lower  denomination  (hr.)  as  fol- 
lows: If  1  day  =  24  hr.,  15  days  will  equal 
15  times  24  hr.,  or  360  hr.,  and  15  days  17  hr. 
will  equal  360  hr.  +  17  hr.,  or  377  hr. 

Next  reduce  377  hr.  to  the  next  lower 
denomination  of  the  scale  (min.)  as  follows: 
If  1  hr.  =  60  min.,  377  hr.  will  equal  377 
times  60  min.,  or  22620  min.,  and  377  hr. 
28  min.  will  equal  22620  min.  +  28  min., 
or  22648  min. 

Lastly,  reduce  22648  min.  to  the  next 
lower  denomination  (sec.)  as  follows:  If 
1  min.  =  60  sec.,  22648  min.  will  equal 
22648  times  60  sec.,  or  1358880  sec.,  the  required  answer. 

RULE.     Multiply  the  highest  given  denomination  by  the  number 
of  units  of  the  next  given  denomination  which  equal  one  unit  of  that 


22648  (377  X  60  +  28) 
X  60  (sec.  in  1  min.) 


1358880  seconds. 


100  COMPOUND    NUMBERS 

highest  denomination,  and  to  the  result  add  the  given  units  of  that 
next  given  denomination. 

Reduce  the  obtained  results  in  a  similar  manner  to  the  next  lower 
given  denomination,  and  continue  this  process  until  all  the  given 
denominations  have  been  considered  and  the  required  lower  denomi- 
nation has  been  obtained. 

NOTE.  It  should  be  remembered  that  a  multiplicand  and  multiplier  are 
not  so  called  because  they  have  a  certain  position  with  respect  to  each  other, 
but  because  they  possess  a  certain  characteristic  relation  to  each  other  [Note 
1,  26].  Thus,  in  the  111.  Ex.,  24,  60,  and  60  are  the  successive  multiplicands 
though  placed  for  convenience  in  positions  usually  occupied  by  multipliers. 

EXAMPLES  FOR  PRACTISE 

Reduce 

1.  £39  18s.  7d.  to  pence 

2.  73  T.  16  cwt.  to  pounds 

3.  16  bu.  3  pk.  5  qt.  to  pints 

4.  $19  and  8  i  to  mills 

5.  3  yr.  216  da.  12  min.  to  seconds 

6.  18  lb.  11  pwt.  to  grains 

7.  £196  16s.  2d.  to  pence 

8.  28  rd.  2  yd.  1  ft.  to  inches 

9.  25  cd.  3  cd.  ft.  to  cubic  feet 

10.  4  A.  25  sq.  rd.  16  sq.  yd.  to  square  feet 

11.  27  mi.  18  rd.  2  yd.  to  inches 

12.  5  T.  12  cwt.  16  Ib.  to  ounces 

13.  3  wk.  4  da.  17  hr.  to  seconds 

14.  35  bu.  2  pk.  3  qt.  to  quarts 

15.  65  gal.  3  qt.  1  pt.  to  gills 

16.  73  A.  75  sq.  rd.  12  sq.  yd.  to  square  inches 

17.  25  cu.  yd.   13  cu.  ft.  to  cubic  inches 

18.  How  much  was  received  for  3  bu.  2  pk.  chestnuts  if  sold 
at  6  cents  per  quart? 

19.  What  is  the  cost  of  1  Ib.  7  oz.  13  pwt.  of  jewelers'  gold 
at  $1.05  per  pennyweight? 

20.  How  much  should  be  received  for  37  gallons  of  cider  if 
retailed  at  5  cents  per  pint? 


COMPOUND    NUMBERS 


101 


REDUCTION  OF  SIMPLE   DENOMINATE  NUMBERS  TO 
COMPOUND   DENOMINATE  NUMBERS 

ILLUSTRATIVE  EXAMPLE 


SOLUTION 


4)3470  gi. 
2)857  _j_  2  gi 

4)433  +  1  pt. 
108  gal.  +  1  qt. 


157.   Reduce  3470  gills  to  the  higher  denominations. 

EXPLANATION.     4  gills  of  the  given  denomination 
(3470  gills)  equal  1  pint,  the  next  higher  denomina- 
tion;   and  3470  gi.  contain  4  gi.  867  times  and  2  gi. 
remaining.     Therefore  3470  gi.  =  867  pt.  and  2  gi. 
2  pt.  of  the  obtained  denomination  (867  pt.) 
equai  i  qt  t>  the  next  higher  denomination;  and  867 
Pt-   contain  2  pt.  433  times  and  1    pt.  remaining. 
Therefore,  867  pt.  =  433  qt.  and  1  pt. 
4  qt.  of  the  last  obtained  denomination  (433  qt.)  equal  1  gal.,  the  next 
higher  denomination;   and  433  qt.  contain  4  qt.  108  times  and  1  qt.  remain- 
ing.    Therefore  433  qt.  =  108  gal.  and  1  qt.;  and  3470  gi.  =  108  gal.  1  qt. 
1  pt.  2  gi. 

RULE.  Reduce  the  given  denomination  to  the  next  higher  denom- 
ination. Continue  this  process  in  regular  order  from  one  obtained 
denomination  to  the  next  higher  until  the  highest  required  denom- 
ination is  reached. 

NOTE,  (a)  Each  successive  divisor  should  be  of  the  same  denomination 
as  its  dividend;  (6)  each  successive  remainder  is  of  the  same  denomination 
as  the  dividend  from  which  it  was  obtained,  and  (c)  each  successive  quotient 
is  of  that  denomination  in  the  scale  which  equals  the  divisor  that  produced  it. 


EXAMPLES  FOR  PRACTISE 


Reduce 

1.  6173  pt.  to  bushels 

2.  3567282  sec.  to  days 

3.  273465  in.  to  miles 

4.  827356  sq.  ft.  to  acres 

5.  23468  far.  to  £'s 

6.  853764  oz.,  Avoir.,  to  tons 

7.  5295  sheets  to  reams 

8.  7835  cu.  ft.  to  cords 

9.  2346  pwt.  to  Ibs,  Troy 
10.  3496  cd.  ft.  to  cords 


Reduce 

11.  7683d.  to  £'s 

12.  53867  mills  to  dollars 

13.  6287  gi.  to  gallons 

14.  2863452  sec.  to  days 

15.  627853  in.  to  miles. 

16.  7456  qt.  to  bushels 

17.  876345  ft.  to  miles 

18.  917345  cu.  in.  to  cu. 

19.  17832d.  to  £'s 

20.  1378  gi.  to  gallons. 


yds. 


102  COMPOUND    NUMBERS 

REDUCTION  OF  DENOMINATE  COMMON  FRACTIONS  TO 

INTEGERS 

ILLUSTRATIVE  EXAMPLE 
158.   Reduce  f  of  a  year  to  integers  of  the  lower  denominations. 

SOLUTIONS  EXPLANATION.  $  of  a  year 

are  equivalent  to  $  of  (times)  365 

4  of  365  da.  =  202  da.  (and  |  da.)  da.  or  202£  da.  Reserving  the 

integral  part  of  this  result  (202 

|  of  24  hr.  =  18  hr.  (and  f  =  f  hr.)     da>)  as  the  first  term  of  the  re- 

.~      .        ,,  quired   answer,   similarly   reduce 

I  of  60  mm.  =  40  mm.  Hence,  the  fractional  part  (|  da  }  to  the 


I  yr.  =  202  da.  18  hr.  40  min.  next  lower  denomination  0*->  *» 

follows: 

|  of  a  day  =  1  of  (times)  24  hr.,  or  18  f  hr.  Reserving  the  integral 
part  of  the  result  (18  hr.)  as  the  second  term  of  the  required  answer,  re- 
duce the  fractional  part  (f  hr.)  to  the  next  lower  denomination  (min.),  as 
follows: 

f  of  an  hour  equals  f  of  (times)  60  min.,  or  40  min.  Therefore  f  of  a  year 
equals  202  da.  18  hr.  40  min. 


EXAMPLES  FOR  PRACTISE 
Reduce  to  integers  of  the  lower  denominations. 

1.  £f          3.    A  gal.        5.  |  yr.  7.  %  yd.          9.  £« 

2.  |  yr.       4.  TV  bu.         6.  |  mi.          8.  A  T.        10.  H  yd. 


REDUCTION  OF  DENOMINATE  DECIMALS  TO  INTEGERS 
ILLUSTRATIVE  EXAMPLE 

159.   Reduce  £.375  to  integers  of  the  lower  denominations. 
SOLUTION 

p  075  EXPLANATION.     .375  of  a   £  =  .375  of   (times)   20s.,  or 

7.5s.     Reserving  the  integral  part  of  this  result  (7s.)  as  the 
first  term  of  the  required  answer,  reduce  the  fractional  part 


S.  7.500  (.5s)  to  the  next  lower  denomination  (d.)  as  follows: 

12  .5s  =  .5  of  (times)  12d.,  or  6d.      Therefore  £.375  =  7s.6d. 

d.1U) 

£.375  =  7s.  6d. 


COMPOUND   NUMBERS  103 

EXAMPLES  FOR  PRACTISE 

Reduce  to  integers  of  the  lower  denominations. 

1.  .1875  gal.  4.  .84375  gal.  7.  .34375  mi. 

2.  .0375  yr.  5.  .876  T.  8.  £.52375 

3.  .546875  bu.  6.  .375  lb.,  Troy  9.  .921875  bu. 

REDUCTION  OF  COMPOUND  INTEGRAL  NUMBERS  TO 
FRACTIONS  OF  ANY  REQUIRED  HIGHER  DENOMINATION 

ILLUSTRATIVE  EXAMPLES 
160.   1.   Reduce  2  pk.  3  qt.  to  a  common  fraction  of  a  bushel. 

SOLUTION  EXPLANATION.  1  pk.  = 

8  qt.,  therefore  2  pk.  3  qt. 

2  pk.  3  qt.  =  (8  qt.  X  2)  +  3,  or  19  qt.  =  2  times  8  qt.  (16  qt.)  +  3 

qt.,  or  19  qt. 

1  bu.  =  4  pk.,  or  4  times  8  qt.  =  32  qt.  1  bu.  =  4  pk.,  and  4  pk. 

o  u      o  i  <»     r     u     u  i  =4  times  8  qt.,  or  32  qt. 

Hence,  2  bu.  3  qt.  =  if  of  a  bushel  If  32  ^  J  l  ^  *,  q{ 

must  equal  A  of  a  bu.,  and  2  pk.  3  qt.,  or  19  qt.,  must  equal  19  times  •??  of 
a  bu.,  or  If  of  a  bu. 

2.   Reduce  17s.  6d.  to  a  decimal  of  a  £  sterling. 

SOLUTION         EXPLANATION.     12d.  =  Is.;   therefore  6d.  must  be  such  a 
12)6  0  decimal  part  of  a  shilling  as  6d.  are  of  12d.,  that  is,  j\  or  .5  of  a 

— »  shilling  [123];    and  17s.  6d.  must  equal  17.5  shillings. 

20s.  =  £1;   and  17.5s.  must  be  such  a  decimal  part  of  a  £ 

20)  17.5          ^  17-5s-  are  of  20s., that  is  ~^T  or  -875  of  a  £. 

£.875  Or,  17s.  6d.  may  be  reduced  to  a  common  fraction  of  a  £ 

as  in  the  preceding  solution,  obtaining  ££,  and  this  result  re- 
duced to  a  decimal  by  123,  obtaining  £.875. 

EXAMPLES  FOR  PRACTISE 

What  fraction  of 

1.  A  £  are  2s.  6d. 

2.  A  dollar  are  12 £  5  mills. 

3.  A  gallon  are  2  qt.  1  pt. 

4.  A  bushel  are  1  pk.  4  qt. 

5.  A  day  are  4  hr.  48  min. 


104  COMPOUND    NUMBERS 

What  decimal  of 

6.  A  gallon  are  3  qt.  1  pt. 

7.  A  £  are  12s.  6d. 

8.  A  yard  are  1  ft.  6  in. 

9.  A  bushel  are  1  pk.  7  qt.  1  pt.  (to  4  places). 
10.  A  day  are  5  hr.  19  min.  (to  3  places). 

ADDITION  OF  COMPOUND  NUMBERS 

ILLUSTRATIVE  EXAMPLE 
161.   Add  3  gal.  2  qt.  1  pt.;  2  gal.  1  pt.;  and  3  qt.  1  pt. 

SOLUTION  EXPLANATION.     So  arrange  the  given  compound  num- 

bers that  their  denominations  which  are  similar  shall  fall 
gal.  qt.  pt.         in  the  same  column  [Prin.  1,  18].      Then  add  the  column 
321         of  the  lowest  given  denomination  (pt.),  obtaining  3  pt.,  or 
2        Q        l          1  qt.  1  pt.     Write  1  pt.  under  the  column  of  pints,  and 
carry  1  qt.  to  the  column  of  quarts  and  add,  obtaining  6  qt., 

or  1  gal.  2  qt.     Write  2  qt.  under  the  column  of  quarts, 

6        2  and  carry  1  gal.  to  the  column  of  gallons  and  add,  obtaining 

6  gal. 

EXAMPLES  FOR  PRACTISE 

123 

da.  hr.  min.  sec.  T.   cwt.  Ib.   oz.  £       s.      d.  far. 

5  13  25  19  45  18  29  9  15  16  8  3 

17  21  18  27  37  3  75  3  7  10  6  1 

12  15  35  40  15  12  34  11  24  3  10  2 

8      7    43     52  7      6    17      6  18    12      93 

4.  AddSbu.  7qt.;  3  bu.  2pk.;  7  bu.  1  pt.;  and  3  pk.  5  qt. 

5.  Add  £35  7s.  6d.;   £53  8d.;   £65  2  far.;   £45  15s. 

6.  Add   16  da.  18  hr.  5  sec. ;    7  da.  3  hr.   48  min. ;     19  da. 
7  hr.  50  sec. ;    12  da.  2  hr.  34  sec. ;    and  2  da.  23  hr. 

7.  Add  19  cu.  yd.  16  cu.  ft.  1217  cu.  in. ;    35  cu.  yd.  24  cu.  ft. 
197  cu.  in.;    2  cu.  yd.   15  cu.  ft.  and  718  cu.  in.;    13  cu.  yd. 
16  cu.  ft. ;    and  25  cu.  yd.  11  cu.  ft.  1235  cu.  in. 

8.  Add  ^  da.;     A  da.;  H  da.;  and  $  da.;  reduced  to  com- 
pound integers. 

9.  Add  .875  cd.  ft.;   .25  cd.;   .125  cd.;  reduced  to  compound 
integers. 


COMPOUND    NUMBERS  105 

SUBTRACTION   OF  COMPOUND   NUMBERS 

ILLUSTRATIVE  EXAMPLE 
162.   Subtract  £15  12s.  3  far.  from  £48  7s.  6d. 

SOLUTION  EXPLANATION.     So  place  the   several   denom- 

inations of  the  subtrahend  that  they  shall  fall 

£       s.     d.     far.  under  sjmilar  denominations  of  the  minuend;  and 

48       7     6       0  commence  the  subtraction  with  the  lowest  de- 

15     12     0       3  nomination. 

As  3  far.  cannot  be  subtracted  from  0  far. 


£32  15s.  5d.  1  far.  borrow  Id.  (=4  far.)  from  the  next  higher  de- 
nomination of  the  minuend  (6d.),  and  then  sub- 
tract 3  far.  from  the  borrowed  4  far.,  obtaining  1  far.  as  the  first  partial 
remainder. 

Next  subtract  the  next  higher  denomination  of  the  subtrahend  (Od.)  from 
the  similar  denomination  of  the  minuend  (6d.  less  the  borrowed  Id.,  or  5d.), 
obtaining  5d.  as  the  second  partial  remainder. 

As  the  next  higher  denomination  of  the  subtrahend  (12s.)  cannot  be  sub- 
tracted from  the  similar  denomination  in  the  minuend  (7s.),  borrow  £1 
(=  20s.)  from  the  £48  of  the  minuend,  and  then  subtract  12s.  from  7s.  in- 
creased by  the  borrowed  20s.,  or  27s.,  obtaining  15s.  as  the  third  partial 
remainder. 

Lastly,  subtract  the  next  higher  denomination  of  the  subtrahend  (£15) 
from  the  similar  denomination  of  the  minuend  (£48  less  the  borrowed  £1, 
or  £47),  obtaining  £32  as  the  final  partial  remainder;  thus  making  the  com- 
plete remainder  £32  15s.  5d.  1  far. 

EXAMPLES  FOR  PRACTISE 

1.  Subtract  5  bu.  1  pk.  1  pt.  from  12  bu.  3  qt. 

2.  Subtract  19  da.  18  hr.  45  min.  from  45  da.  12  hr. 

3.  Subtract  25  cd.  95  cu.  ft.  from  42  cd.  37  cu.  ft. 

4.  Subtract  167  gal.  3  qt.  2  gi.  from  258  gal.  2  qt.  1  pt. 

5.  Subtract  £235  17s.  6d.  from  £452  12s.  9d. 

6.  Subtract  15  A.  90  sq.  rd.  18  sq.  yd.  and  5  sq.  ft.  from 
23  A.  73  sq.  rd.  24  sq.  ft. 

7.  A  grocer  sold  27  gal.  3  qt.  1  pt.  from  a  barrel  of  molasses 
containing  39  gal.   1  qt.     How  much  molasses  remained  in  the 
barrel? 

8.  A   merchant   in   London   owed    £285  2s  6d   and   subse- 
quently made  a  payment  of  £97  12s  9d.     How  much  of  the 
debt  remained  unpaid? 


106  COMPOUND    NUMBERS 

COMPOUND  SUBTRACTION  OF  DATES 
ILLUSTRATIVE  EXAMPLE 

163.  Find  the  interval  between  Oct.  25,  1854,  and  July  16, 
1911. 

SOLUTION  EXPLANATION.     Take  the  later  date  (July 

16,  1911)  as  the  minuend  because  it  expresses 

yr.  mo.       da.  a  greater  period  of  time,  denoting  the  1911th 

1911  7          16  year  of  the  Christian  era,  the  7th  month  [Note 

1854          10          25  2»  153^»  and  the  16th  dav»'    underneath  which 

— ^ft —    — ft 9?  r\          wr*te  tne  earlier  date  which   expresses  a  less 

00  yr.     8  mo.  21  da.      period  of  time,  denoting  the  1854th  year  of  the 
Christian  era,  the  10th  month,  and  the  25th 
day.     Subtract  as  shown  in  111.  Ex.  162. 

NOTE  1.  In  obtaining  the  interval  between  two  dates  by  compound 
subtraction,  it  is  customary  to  consider  any  borrowed  month  as  equal  to  30 
days. 

NOTE  2.  If  both  the  subtrahend  and  minuend  express  the  last  days  of 
their  respective  months,  and  the  subtrahend  month  is  the  longer,  as  in  Ex.  7 
and  9,  following,  the  difference  between  them  will  be  0;  but  if  the  subtrahend 
month  is  the  shorter  as  in  Ex.  10,  following,  the  difference  is  found  in  the  usual 
manner.  The  courts  have  decided  when  a  male  child  is  born  on  Feb.  29th  of 
a  leap  year,  that  he  has  reached  his  majority  on  Feb.  28th  of  the  21st  year 
thereafter;  and  when  a  note  falls  due  on  the  29th,  30th,  or  31st  of  a  month 
which  contains  less  than  that  number  of  days,  that  it  is  payable  on  the  last 
day  of  such  a  month.  Thus,  3  months  after  Aug.  31  =  Nov.  30,  and  con- 
versely from  Aug.  31  to  Nov.  30  must  equal  3  months. 

EXAMPLES  FOR  PRACTISE 
By  compound  subtraction  find  the  interval  from 

1.  Mch.  16,  1892  to  Dec.  3,  1911. 

2.  Oct.  28,  1886  to  May  4,  1912. 

3.  July  2,  1875  to  Feb.  12,  1912. 

4.  Dec.  18,  1895  to  Aug.  9,  1911. 
6.  Nov.  30,  1845  to  Feb.  8,  1910. 

6.  Sept.  19,  1902  to  July  31,  1910. 

7.  Aug.  31,  1897  to  June  30,  1911.  [Note  2] 

8.  May  31,  1875  to  Jan.  31,  1909. 

9.  Feb.  29,  1892  to  Feb.  28,  1910. 
10.  Feb.  28,  1887  to  Feb.  29,  1912.  * 


COMPOUND    NUMBERS  107 

TO  FIND  THE  EXACT  NUMBER  OF  DAYS 
BETWEEN  TWO  DATES 

ILLUSTRATIVE  EXAMPLE 

164.  How  many  days  after  Oct.  13,  1911,  is  Apr.  24,  1912? 

SOLUTION 

•    j        t  r\  A  u  EXPLANATION.     Subtract  the  initial 

18  da.,  remainder  of  October  date  (Qct   13)  from  ^  full  number  of 

30  da.,  in  November  days  m  that  month  (31))  obtaining  18 

31  da.,  in  December  days  as  the  included  remainder  of  Oct., 
31  da.,  in  January  under  which  place  in  regular  order  the 
29  da     in  February  number  of  days  in  each  of  the  succeeding 

«1    ,  A/TV,  ^U^  months  of  the  interval,  and  the  in- 

da.,  in  Marcn  duded  dayg  of  the  final  month  (24). 

24  da.,  included  in  April  g^  M}  obtaining  194  days. 

194  da.,  required  interval 

EXAMPLES  FOR  PRACTISE 

Find  the  exact  number  of  days  from 

1.  Jan.  16,  1911  to  Nov.  12,  1911. 

2.  July  23,  1912  to  Mch.  8,  1913. 

3.  Sept.  3,  1909  to  May  31,  1910. 

4.  Feb.  16,  1910  to  Oct.  6,  1910. 

5.  Dec.  31,  1911  to  Aug.  12,  1912. 

6.  Nov.  16,  1910  to  Feb.  3,  1911. 

7.  Mch.  26,  1911  to  Dec.  20,  1911. 

8.  Apr.  30,  1909  to  Jan.  14,  1910. 

9.  June  16,  1911  to  Apr.  8,  1912. 

10.  May  19,  1910  to  May  28,  1910. 

11.  Aug.  27,  1911  to  Mch.  5,  1912. 

12.  Oct.  1,  1907  to  July  17,  1908. 

MULTIPLICATION  OF  COMPOUND  NUMBERS 
ILLUSTRATIVE  EXAMPLE 

165.  Multiply  28  da.  16  hr.  34  min.  by  6. 

SOLUTION  EXPLANATION.     Commencing    with    the 

da         hr        min  lowest  denomination   (34  min.)   successively 

'  '  '  multiply  each  denomination  of  the  multipli- 

cand by  the  multiplier,  as  follows: 

5.  6   times   34   min.  =  204  min.,    or   3   hr. 

172  da    3  hr.     24  min.          24  min.     Write  24  min.  as  a  partial  product, 


108  COMPOUND    NUMBERS 

and  carry  3  hr.  to  6  times  16  hr.,  or  96  hr.,  obtaining  99  hr.,  or  4  da.  3  hr. 
Write  3  hr.  as  a  second  partial  product,  and  carry  4  da.  to  6  times  28  da., 
or  168  da.,  obtaining  172  da.  as  a  third  and  final  partial  product;  thus  mak- 
ing the  complete  product  equal  172  da.  3  hr.  24  min. 


EXAMPLES  FOR  PRACTISE 

Multiply  Multiply 

1.  5  gal.  2  qt.  1  pt.  by  8  7.    17  mi.  48  rd.  4  yd.  2  ft.  by  73 

2.  £24  3s.  5d.  by  7  8.   53  yr.  7  mo.  23  da.  by  85 

3.  16  da.  21  hr.  15  min.  by  9      9.    78  bu.  3  pk.  2  qt.  1  pt.  by  147 

4.  46  Ib.  8  oz.  12  pwt.  by  6       10.   47  T.  15  cwt.  73  Ib.  by  524 

5.  53  bu.  2  pk.  5  qt.  by  12         11.    16  da.  15  hr.  49  min.  by  97 

6.  £273  16s.  7d.  3  far.  by  97      12.    9  gal.  3qt.  1  pt.  by  276 

13.  The  distance  between  two  towns  is  42  miles.  If  a  man 
walk  from  one  towards  the  other  for  8  hours  at  an  average  speed 
of  3  miles  and  192  rods  per  hour,  how  far  will  he  then  be  from 
the  other  town? 

DIVISION  OF  COMPOUND  NUMBERS  BY  ABSTRACT 

NUMBERS 

ILLUSTRATIVE  EXAMPLES 
166.   1.   Divide  35  Ib.  10.  oz.  18  pwt.  by  8. 

SOLUTION  EXPLANATION.     Commencing  with 

H,    irv  the  highest  denomination  (35   Ib.)  suc- 

8)35  Ib.  10  oz.  18  pwt.  cessively  divide  each    denomination  of 

4  Ib.     5  OZ.  17  pwt.  6  gr.     the  dividend  by  the  divisor  as  follows: 

35  Ib.  -r-  8  =  4  Ib.  and  an  un- 
divided remainder  of  3  Ib.  Write  4  Ib.  as  a  partial  quotient;  and  carry  the 
undivided  remainder  (3  Ib.  or  36  oz.)  to  the  next  lower  denomination  of  the 
dividend  (10  oz.),  obtaining  46  oz.  as  the  second  partial  dividend. 

46  oz.  -5-  8  =  5  oz.  and  an  undivided  remainder  of  6  oz.  Write  5  oz.  as 
the  second  part  of  the  quotient;  and  carry  the  undivided  remainder  (6  oz.  or 
120  pwt.)  to  the  next  lower  denomination  of  the  dividend  (18  pwt.),  obtaining 
138  pwt.  as  the  third  partial  dividend. 

138  pwt.  -f-  8  =  17  pwt.  and  an  undivided  remainder  of  2  pwt.  Write 
17  pwt.  as  the  third  part  of  the  quotient;  and  reduce  the  undivided  remainder 
(2  pwt.)  to  the  next  lower  denomination  of  the  Troy  scale  (2  pwt.  =  48  gr.). 

48  gr.  -T-  8  =  6  gr.  and  no  undivided  remainder.  Write  6  gr.  as  the  fourth 
and  final  part  of  the  quotient.  These  several  partial  quotients  combined 
make  the  complete  quotient  4  Ib.  5  oz.  17  pwt.  6  gr. 


COMPOUND    NUMBERS  109 

2.  Divide  £249  10s.  5d.  by  29. 

SOLUTION 

29) £249  10s.  5d.(£8 
232 

£17  remainder 

X  20  (and  add  10s.)  EXPLANATION.    When  the  divisor 

is  over  12,  and  not  a  composite  num- 

^yjoous.^izs.  ber,    long    division    may  be  advan- 
tageously employed. 

gQ  The    processes    in    the    present 

ro  example  are  similar  to  those  of  the 

—  preceding    example;   but  here    they 

2s.  remainder  are   written    because   they    are    too 

X  12  (and  add  5d.)  unwieldy  to  be  carried  mentally. 

29)29  d.(ld. 

29 
Complete  quotient,  £8  12s.  Id. 

EXAMPLES  FOR  PRACTISE 
Divide  Divide 

1.  39  gal.  1  qt.  1  pt.  by  5.  6.  £189  9s.  4d.  by  56  (=  8  X  7) 

2.  78  yd.  2  ft.  8  in.  by  8  7.  105  T.  4  cwt.  38  Ib.  by  27  (9X3) 

3.  £225  18s.  6d.  by  7  8.  369  gal.  1  pt.  by  47 

4.  155  cd.  50  cu.  ft.  by  9  9.  277  yr.  4  mo.  24  da.  by  73 

5.  287  bu.  2  pk.  4  qt.  by  12  10.  930°  39'  55"  by  269 

DIVISION  OF  COMPOUND  NUMBERS  BY  COMPOUND 

NUMBERS 
ILLUSTRATIVE  EXAMPLE 

167.   Divide  907  yd.  2  ft.  6  in.  by  34  yd.  2  ft.  9  in. 

SOLUTION 

EXPLANATION.  Reduce  both 

K)7  yd.  2  ft.  6  in.  =  32682  in.        compound   numbers   to   simple 
34  yd.  2  ft.  9  in.    =  1257  in.          denominate     numbers     of     the 

lowest  denomination  expressed 

1257)32f  32(26  times  in   either)   obtaining  32682   in. 

2514  as  the  equivalent  simple  divi- 

7542  dend,  and  1257  in.  as  the  equiv- 

alent   simple    divisor.     Divide 
in  the  usual  manner. 


110  REVIEW   OF   COMPOUND    NUMBERS 

EXAMPLES  FOR  PRACTISE 

1.  Divide  £1000  8s.  4d.  by  £35  14s.  7d. 

2.  Divide  764  bu.  by  5  bu.  3  pk.  7  qt. 

3.  Divide  1172  gal.  2  qt.  1  pt.  by  19  gal.  3  qt.  1  pt. 

4.  Divide  53  da.  11  hr.  56  min.  54  sec.  by  17  hr.  35  min.  18  sec. 

5.  Divide  183  Ib.  3  oz.  12  pwt.  by  3  Ib.  9  oz.  16  pwt.  12  gr. 

REVIEW  OF  COMPOUND  NUMBERS 

168.  1.  From  a  field  of  wheat  containing  78  A.,  was  gathered 
an  average  crop  of  28  bu.  1  pk.  3  qt.  per  acre.  How  much  wheat 
did  the  field  produce? 

2.  How  many  spoons  weighing  1  oz.  17  pwt.  can  be  made 
from  3  Ib.  8  oz.  8  pwt  of  silver? 

3.  A  tenant  rented  a  house  on  March  18,  1909,  at  $22.50  per 
month,  and  occupied  it  until  February  18,  1912.     What  was  the 
total  rent  paid  during  his  tenancy? 

4.  A  man  contracted  a  debt  on  June  16,  1910,  and  discharged 
it  on  April  23, 1911.     How  many  days  did  the  debt  remain  unpaid? 

5.  A  coal  dealer  bought  58240  Ib.  of  coal  at  $5.20  per  long  ton 
and  sold  it  at  $6.10  per  short  ton.     What  was  his  total  gain? 

6.  A  customer  bought  a  piece  of  goods  at  2s.  6d.  per  yard,  and 
paid  £5  7s.  6d.     How  many  yards  did  the  piece  contain? 

7.  A  grocer  bought  3  bu.  2  pk.  7  qt.  of  peanuts  and  retailed 
his  purchase  at  5^  per  pint.     How  much  did  he  receive? 

8.  What  is  the  difference  between  15  Ib.  Avoir.,  and  25  Ib. 
Troy  expressed  in  denominations  of  Troy  Weight? 

9.  A  charity  organization  divided  51  bu.  1  pk.  7  qt.  of  pota- 
toes equally  between  27  needy  families.     How  much  did  each 
family  receive? 

10.  A  gardener  divided  H  of  an  acre  of  ground  into  12  equal 
vegetable  plats.     How  much  space  did  each  plat  occupy,  expressed 
integrally? 

11.  An  importer  paid  £253  5s  7d.  for  one  invoice  of  merchan- 
dise, £75  18s.  6d.  for  a  second  invoice,  £148  12s.  9d.  for  a  third, 
£62  9s.  3d.  for  a  fourth,  and  £197  17s.  3d.  for  a  fifth.     What  was 
his  total  payment  for  the  five  invoices? 

12.  A  grocer  sold  18  gal.  3  qt.  1  pt.  2  gi.  of  vinegar  from  a 


THE    METRIC    SYSTEM  111 

barrel  containing  42  gal.  1  qt.     How  much  vinegar  remained  in 
the  barrel? 

13.  A  pedestrian  walked  7  hours  per  day  at  an  average  speed 
of  4  mi.  37  rd.  3  yd.  per  hour.     What  distance  had  he  traveled  at 
the  end  of  the  ninth  day's  walk? 

14.  How  much  will  be  received  for  a  barrel  of  cider  contain- 
ing 42.375  gallons,  if  retailed  at  3^f  per  pint? 

15.  How  many  barrels  will  be  needed  to  ship  79  bu.  of  pota- 
toes, if  the  average  capacity  per  barrel  is  2  bu.  1  pk.  7  qt? 

16.  What  is  the  value  in  United  States  money  of  873.45  marks 
(Germany)?     17.   Of  728.75  crowns  (Sweden)?     18.   Of  £673  2s. 
6d.  (Great  Britain)?     19.   Of  3278  lira  (Italy)?     20.   Of  348.25 
yen  (Japan)?    21.   Of  758.35  francs  (France)? 

22.  Reduce  $675.80  to  French  francs.  23.  $528.36  to  £'s 
sterling.  24.  $912.50  to  Russian  rubles.  25.  $248.75  to  Spanish 
peseta. 

THE  METRIC  SYSTEM 

169.  INTRODUCTION.   The  metric  system  is  the  expression  of 
compound  denominate  numbers  upon  the  decimal  scale;   and  by 
its  use  the  convenient  processes  incident  to  operations  upon  simple 
integral  quantities  and  upon  decimal  fractions  can  be  substituted 
in  the  place  of  the  inconvenient  methods  necessary  for  operating 
upon  compound  integral  quantities  and  upon  common  fractions 
which  are  expressed  upon  an  arbitrary  variable  scale.     The  system 
derives  its  name  from  the  meter  (39.37079  inches  long)  which  is 
the  established  standard  not  only  for  measuring  distances,  but 
also  for  determining  the  size  of  the  other  standards  for  weights, 
measures,  surfaces,  and  volumes,  by  expressing  them  in  decimal 
multiples  or  aliquots  of  the  meter. 

170.  The   higher   denominations   of   the   metric   system   are 
denoted  by  prefixing  the  Greek    numerals    deka   (10   standard 
units),  hekto  (100  standard  units),  kilo  (1000  standard  units),  and 
myria  (10,000  standard  units). 

Thus,  1  dekameter  means  10  meters;  5  hektometers,  500  meters;  7  kilo- 
meters, 7000  meters;  and  9  myriameters,  90000  meters.  The  metric  com- 
pound 5  hektometers  9  dekameters  and  7  meters,  is  written  597  M.;  the 
meters  (7)  occupying  the  units'  order,  the  dekameters  (9)  the  tens'  order,  the 
hektometers  (5)  the  hundreds'  order,  in  accordance  with  the  signification  of 
their  respective  prefixes. 


112  THE    METRIC    SYSTEM 

171.  The  lower  denominations  of  the  metric  system  are  de- 
noted by  prefixing  the  Latin  numerals  deci  (one-tenth  of  one 
standard  unit),  centi  (one-hundredth  of  one  standard  unit),  milli 
(one-thousandth  of  one  standard  unit). 

Thus,  1  decimeter  means  one-tenth  of  a  meter;  3  centimeters  mean  3 
hundred ths  of  a  meter;  4  millimeters  mean  4  thousandths  of  a  meter.  The 
metric  compound  2  decimeters,  4  centimeters,  3  millimeters  is  written  .243  M., 
in  accordance  with  the  signification  of  their  respective  prefixes;  and  is  usually 
read  243  thousandths  of  a  meter,  or  243  millimeters. 

172.  The  meter  (=  39.37079  inches)  is  the  metric  standard 
for  measuring  lengths,  widths,  heights,  depths  and  distances. 

NOTE  1.  The  meter  (pronounced  mee'ter)  is  used  by  merchants  in  meas- 
uring dry  goods,  and  by  others  in  measuring  short  distances;  the  millimeter 
being  used  in  measuring  minute  distances;  and  the  kilometer  in  measuring 
long  distances. 

NOTE  2.  The  higher  and  lower  denominations  of  the  meter  are  expressed 
by  the  use  of  the  prefixes  mentioned  in  170  and  171. 

NOTE  3.  Reductions  from  the  common  denominations  of  length  to  the 
metric,  or  the  reverse,  may  be  made  by  the  use  of  the  following 


EQUIVALENTS    OF    LINEAR    MEASURE 

(Common  to  Metric)  (Metric  to  Common) 

1  inch  =  2.54001  centimeters  1  centimeter  =  .3937  of  an  inch 

1  foot  =  .304801  of  a  meter  1  decimeter  =  .32818  of  a  foot 

1  yard  =  .914402  of  a  meter  1  meter  =  1.093611  yards 

1  rod  =  5.029211  meters  1  dekameter  =  1.98838  rods 

1  mile  =  1.60935  kilometers  1  kilometer  =  .62137  of  a  mile 

173.  The  gram  (=  .03527  oz.,  Avoir.,  or  15.432  gr.,  Troy)  is 
the  metric  standard  for  weights.  It  is  the  weight  of  a  cubic 
volume  of  distilled  water  each  edge  of  which  is  .01  of  a  meter. 

NOTE  1.  The  higher  and  lower  denominations  of  the  gram  are  expressed 
by  the  use  of  the  same  prefixes  with  the  same  signification  as  those  of  the  meter 
[170,  171].  Thus,  8  hektograms  mean  800  grams;  4  milligrams  mean  4  thou- 
sandths of  a  gram;  etc. 

NOTE  2.  The  gram  is  used  to  express  the  weight  of  minute  quantities,  as 
jewels,  medicines,  etc.;  and  the  kilogram  (usually  abbreviated  to  kilo)  to 
express  ordinary  weights. 


THE    METRIC    SYSTEM  113 

EQUIVALENTS    OF    WEIGHT 

(Common  to  Metric)  (Metric  to  Common) 

1  oz.,  Avoir.  =  28.3494  grams  1  gram  =  .03527  oz.,  Avoir. 

1  oz.,  Troy  =  31.104  grains  1  gram  =  .03215  oz.,  Troy 

1  lb.,  Avoir.  =  .45359  of  a  kilo.  1  hektogram  =  3.527  oz.,  Avoir. 

1  lb.,  Troy  =  .37324  of  a  kilo.  1  kilogram  =  2.20462  lb.,  Avoir. 

174.  The  liter   (=  1.0567    liquid  quarts,   or    .908  of    a  dry 
quart)  is  the  metric  standard  for  capacities.     It  is  the  capacity 
of  a  cubic  measure  each  edge  of  which  is  one-tenth  of  a  meter. 
The  liter  is  used  to  measure  not  only  liquids,  but  also  salt,  grains, 
fruits,  vegetables,  etc.     It  is  pronounced  "lee'ter." 

NOTE  1.  The  higher  and  lower  denominations  of  the  liter  are  denoted 
by  the  prefixes  mentioned  in  170,  171. 

NOTE  2.  Of  liquids,  the  dekaliter  is  used  to  express  large  quantities,  the 
liter  moderate  quantities,  and  the  centiliter  or  milliliter  minute  quantities. 
Of  grains,  fruits,  vegetables,  the  hektoliter,  like  the  bushel,  is  used  to  express 
large  quantities  (wholesale);  and  the  dekaliter  or  liter,  small  quantities  (retail). 

EQUIVALENTS    OF    CAPACITY 

(Common  to  Metric)  (Metric  to  Common) 

1  liquid  quart  =  .94636  of  a  liter  1  liter  =  1.05668  liquid  qts. 

1  dry  quart  =  1.1012  liters  1  liter  =  .9081  of  a  dry  qt. 

1  liquid  gallon  =  .37854  of  a  dekaliter  1  dekaliter  =  2.6417  liquid  gal. 

1  peck  =  .881  of  a  dekaliter  1  dekaliter  =  1.1351  pecks 

1  bushel  =  .35239  of  a  hektoliter  1  hektoliter  =  2.8375  bu. 

175.  The  ar  (  =  119.6034  square  yards)  is  the  metric  stan- 
dard for  measuring  land.     It  is  a  square  each  side  of  which  is  10 
meters,  and  is  therefore  equal  to  100  square  meters.     It  is  pro- 
nounced "air." 

Scale:     100  centars  =  1  ar;   100  ars  =  1  hektar. 

176.  The  square  meter  (=  1.196  square  yards)  is  the  metric 
standard  for  measuring  surfaces  other  than  those  of  land.     It  is 
a  square  each  side  of  which  is  1  meter. 

Scale:  100  sq.  centimeters  =  1  sq.  decimeter;  100  sq.  deci- 
meters =  1  sq.  meter;  100  sq.  meters  =  1  sq.  dekameter. 


114  THE    METRIC    SYSTEM 

NOTE.  As  in  metric  square  measures  100  units  of  any  order  make  one 
unit  of  the  next  higher  order,  it  is  necessary  to  give  each  order,  except  the  high- 
est, two  places  in  expressing  either  land  surfaces  [175]  or  ordinary  surfaces 
[176].  Thus,  19  sq.  dekameters  5  sq.  meters  49  sq.  decimeters  9  sq.  centi- 
meters should  be  written  1905.4909  M.;  and  2  hektars  7  ars  5  centars  of  land 
are  written  207.05  ars. 

EQUIVALENTS  OF  SQUARE  MEASURE 

(Common  to  Metric)  (Metric  to  Common) 

1  sq.  inch  =  6.4516  sq.  centimeters  1  sq.  centimeter  =  .155  of  a  sq.  in. 

1  sq.  foot  =  .0929  of  a  sq.  meter  1  sq.  decimeter  =  .1076  of  a  sq.  ft. 

1  sq.  yard  =  .8361  of  a  sq.  meter  1  sq.  meter  =  1.1960  sq.  yd. 

1  sq.  rod  =  25.293  sq.  meters  1  ar  =  3.954  sq.  rods 

1  acre  =  40.469  ars  1  hektar  =  2.47104  acres 

1  sq.  mile  =  259  hektars  1  sq.  kilometer  =  .3861  of  a  sq.  mi. 

177.  The  ster  (=  .2759  of  a  cord)  is  the  metric  standard  for 
measuring  wood.     It  is  a  cube  each  edge  of  which  is  1  meter. 
It  is  pronounced  "stair." 

Scale:     10  decisters  =  1  ster;  10  sters  =  1  dekaster. 

178.  The  cubic  meter  is  the  metric  standard  for  measuring 
solids  other  than  wood.     Like  the  Ster  [177],  it  is  a  cube  each  edge 
of  which  is  1  meter. 

Scale:  1000  cubic  millimeters  =  1  cubic  centimeter;  1000 
cubic  centimeters  =  1  cubic  decimeter;  1000  cubic  decimeters  = 
1  cubic  meter. 

NOTE.  As  in  metric  cubic  measure  1000  units  of  any  order  make  1  unit 
of  the  next  higher  order,  it  is  necessary  to  allot  three  places  to  each  order  except 
the  highest  in  expressing  solids  other  than  wood.  In  wood  measure  the  scale 
is  10,  and  therefore  only  one  place  is  required  to  express  each  order. 

EQUIVALENTS    OF  CUBIC    MEASURE 

(Common  to  Metric)  (Metric  to  Common) 

1  cu.  inch  =  16.387  cu.  centimeters  1  cu.  centimeter  =  .061  of  a  cu.  in. 

1  cu.  foot  =  28.317  cu.  decimeters  1  cu.  decimeter  =  .0353  of  a  cu.  ft. 

1  cu.  yard  =  .7645  of  a  cu.  meter  1  cu.  meter  =  1.308  cu.  yards 

1  cord  =  3.624  sters  1  ster  =  .2759  of  a  cord 

179.  Reductions  from  one  metric  denomination  to  another  are 
effected  by  simply  placing  a  decimal  point  at  the  right  of  the 
required  denomination,  removing  the  original  decimal  point,  and 


THE    METRIC    SYSTEM  115 

making  the  corresponding  change  in  the  abbreviation  to  that  of 
the  required  denomination. 

Thus,  to  reduce  to  a  higher  metric  denomination,  remove  the  decimal 
point  one  place  to  the  left  if  the  scale  of  reduction  is  10,  two  places  to  the  left  if 
the  scale  is  100,  or  three  places  to  the  left  if  the  scale  is  1000,  for  each  interme- 
diate denomination  until  the  required  higher  denomination  is  reached,  prefix- 
ing O's  if  necessary;  and  to  reduce  to  a  lower  metric  denomination,  remove  the 
decimal  point  one,  two,  or  three  places  to  the  right,  according  to  the  scale 
used,  for  each  intermediate  denomination  until  the  required  lower  denomina- 
tion is  reached,  annexing  O's  if  necessary. 

NOTE  1.  When  a  metric  quantity  is  expressed  in  figures,  intermediate 
vacant  orders  must  also  be  expressed  by  writing  as  many  O's  for  each  inter- 
mediate vacant  order  as  there  are  ciphers  in  its  scale. 

NOTE  2.  Metric  denominations  are  abbreviated  by  writing  the  initial 
letter  of  the  prefix  followed  by  the  initial  letter  of  the  standard  unit;  and  by 
commencing  the  abbreviation  with  a  capital  letter  when  it  expresses  a  higher 
denomination  than  the  standard  unit  [170],  and  with  a  "small"  letter  when  it 
expresses  a  lower  denomination  than  the  standard  unit  [171]. 

180.  Metric  quantities  are  added,  subtracted,  multiplied,  and 
divided  as  ordinary  decimals. 

NOTE.  Before  adding  or  subtracting,  it  is  necessary  that  the  several 
metric  quantities  should  be  reduced  to  the  same  denomination  [179],  that  is, 
that  the  decimal  point  of  each  should  be  at  the  right  of  the  same  denomination, 
thus  throwing  like  denominations  in  the  same  column  [Prin.  1,  18]. 


EXAMPLES  FOR  PRACTISE 
Reduce  Reduce 

1.  4613.48  Sq.  M.  to  sq.  dm.         8.   268  Km.  to  miles. 

2.  346834.57  dg.  to  Dg.  9.    56  Dl.  of  vinegar  to  gal. 

3.  46.8173  Hg.  to  dg.  10.   475  HI.  of  wheat  to  bush. 

4.  567.8  ds.  to  S.  11.   468  gal.  wine  to  Dl. 

5.  196538.24  ca.  to  Ha.  12.    15.2  Ib.  Troy,  to  dg. 

6.  28.246  Km.  to  mm.  13.   75  yd.  cloth  to  M. 

7.  341728.3  cl.  to  Dl.  14.   347.48  sq.  yd.  to  Sq.  M. 

15.  Add  468.75  M.,  3.567  Dm.,  and  6178.35  cm. 

16.  Subtract  46.8  HI.  from  58722.348  dl. 

17.  Find  the  cost  of  9348.56  Kg.  of  sugar  at  11  cents  per  kilo. 

18.  Find  the  cost  of  735.48  M.  cloth  at  5.37  francs  per  M. 


116  COMPUTATION    OF    SURFACES 

COMPUTATION    OF    SURFACES 

181.  A  surface  is  the  outside  or  visible  part  of  a  solid  body, 
without  reference  to  its  depth. 

NOTE.  A  plane  surface  is  one  which  is  perfectly  flat,  having  no  undula- 
tions. 

182.  A  rectangular    surface  is  a  plane  surface  bounded  by  four 
straight  sides,  each  perpendicular  to  its  adjacent  side. 

183.  A  square  is  a  rectangular  surface  bounded  by  four  equal 
sides. 

Thus,  a  square  foot  is  a  rectangular  surface  each  side  of  which  is  one  foot; 
a  square  yard  is  a  rectangular  surface  each  side  of  which  is  one  yard;  etc. 

184.  The  area  of  a  rectangular  surface  is  the  amount  of  space 
included  within  its  four  sides,  expressed  in  square  units  of  any 
convenient  denomination. 

Thus,  to  find  the  area  of  a  rectangular  surface  1  yd.  1  ft.  long  and  1  yd. 
wide,  a  convenient  measuring  unit  is  one  square  foot.  The  area  can  then  be 
found  as  follows : 

EXPLANATION.     Fig.  1  is  the 

f  FT  3    Fr  3  rr  measuring    unit,    each    side    of 

which  is  1  foot.     Fig.  2  shows 


^|iso/r[         »Z|   3|sq.|pT.| 

FIG.  I  FIG. 2 


12 


SQ 


FT. 


the  number  of  these  measuring 
squares  in  1  row  (3  sq.  ft.). 
Fig.  3  shows  the  number  of  rows 
(4) .  Therefore,  4  times  3  sq.  ft., 
or  12  sq.  ft.  is  the  area  of  a  rec- 
FIG.  3  tangular  surface  which  is  1  yd. 
1  ft.  long,  and  1  yd.  wide. 

THE  LENGTH  AND  BREADTH  BEING  GIVEN 
TO  FIND  THE  AREA 

185.   How  many  sq.  yd.  in  a  floor  18  ft.  long  and  14  ft.  6  in. 

wide? 

EXPLANATION.  A  floor  18  ft. 
long  and  only  one  foot  wide  con- 
tains 18  sq.  ft.;  therefore  a  floor 

18  sq.  ft.  X  14  J  =  261  sq.  ft.          18  ft.  long  and  14£  ft.  wide  must 
261  —  9  =  29  SQ    yd  contain  14|  times  18 sq.ft., or 261 

sq.  ft.,  which,  when  reduced  to 
sq.  yd.  by  167,  equal  29  sq.  yd. 


COMPUTATION    OF    SURFACES  117 

SUMMARY,  (a)  The  number  of  square  measuring  units  in  one 
row  (lying  adjacent  to  one  dimension)  is  the  multiplicand,  (b)  The 
number  of  rows  (as  indicated  by  the  number  of  the  same  square 
measuring  units  lying  adjacent  to  the  other  dimension)  is  the 
multiplier,  (c)  The  area  (the  number  of  square  measuring  units 
in  all  the  rows)  is  the  product. 

NOTE.  Before  finding  any  required  area,  reduce  the  given  dimensions  to 
the  same  denomination  if  not  already  so  expressed. 

EXAMPLES  FOR  PRACTISE 
Find  the  area  of  a  surface  which  is 

1.  32  ft.  long  and  24  ft.  wide. 

2.  56  yd.  long  and  18  ft.  wide. 

3.  7  yd.  long  and  2  ft.  6  in.  wide. 

4.  18  ft.  6  in.  long  and  12  ft.  wide. 

6.  34  yd.  2  ft.  long  and  18  yd.  1  ft.  wide. 

6.  35  rods  long  and  25  yd.  2  ft.  wide. 

7.  173  yd.  1  ft.  long  and  98  yd.  2  ft.  wide. 

8.  18  rd.  2  yd.  long  and  12  rd.  4  yd.  1  ft.  wide. 

9.  3  mi.  190  rd.  long  and  2  mi.  75  rd.  wide. 

10.  15  mi.  long  and  7  mi.  200  rd.  wide.  • 

11.  3  yd.  2  ft.  9  in.  long  and  2  yd.  1  ft.  6  in.  wide. 

12.  8  rd.  4  yd.  long  and  5  rd.  3  yd.  2  ft.  wide. 

13.  How  many  square  yards  in  a  ceiling  28  ft.  9  in.  long  and 
20  ft.  6  in.  wide? 

14.  How  many  square  yards  in  the  walls  of  a  room  30  ft.  long, 
20  ft.  wide,  and  8  ft.  6  in.  high? 

15.  How  many  acres  does  a  rectangular  field  contain  which  is 
350  yd.  2  ft.  long  and  235  yd.  1  ft.  6  in.  wide? 

16.  How  many  sq.  ft.  in  the  outer  surface  of  a  packing  case 
which  is  3  ft.  7  in.  long,  2  ft.  9  in.  wide,  and  2  ft.  3  in.  high? 

17.  How  many  sq.  ft.  of  surface  are  in  a  blackboard  9  ft.  6  in. 
long  and  4  ft.  wide? 

18.  What  is  the  cost  of  paving  a  cellar  floor  with  cement  at  35^ 
per  sq.  yd.,  if  the  cellar  is  30  ft.  long  and  18  ft.  wide? 

19.  What  is  the  cost  of  paving  a  sidewalk  18  ft.  long  and  12  ft. 
wide  at  25  cents  per  square  foot? 


118  COMPUTATION    OF    SURFACES 

TO  FIND  ONE  DIMENSION,   WHEN  THE  AREA  AND  THE 
OTHER  DIMENSION  ARE  GIVEN 

ILLUSTRATIVE  EXAMPLE 

186.  If  a  rectangular  field  contains  40  rods  and  is  22  yards 
wide,  what  is  its  length? 

EXPLANATION.  The  area  (40  rods) 
is  the  given  product  [Summary,  c, 

40  rods=  1210  sq.  yd.,  area        185];    and  22  (times  the  square  yards 

adjoining  the  required  width)  is  the 

1210  -T-  22  =  55  yd.,  length        given  multiplier  factor  of  that  prod- 
uct    [Summary,     b,     186].     Hence, 

complete  the  relation  of  these  terms  of  multiplication  by  dividing  the  given 
product  (40  rods,  or  1210  sq.  yd.)  by  its  given  multiplier  factor  (22),  obtaining 
55  sq.  yd.  (adjoining  the  required  width)  as  its  required  multiplicand  fac- 
tor [Summary,  a,  186].  As  only  one  linear  yard  of  each  of  these  55  sq.  yd. 
is  in  immediate  contact  with  the  required  width,  the  width  must  be  55  linear 
yards. 

NOTE.  Before  division,  the  given  area  should  be  made  to  express  square 
units,  and  the  given  dimension  linear  units,  of  the  same  denomination  as  that 
of  the  required  dimension. 

EXAMPLES  FOR  PRACTISE 

1.  If  the  area  of  a  floor  is  35  sq.  yd.  8  sq.  ft.,  and  its  length  is 
6  yd.  1  ft.,  what  is  its  breadth? 

2.  If  a  rectangular  field  contains  74  acres  and  is  880  yd.  long, 
how  many  rods  is  its  breadth? 

3.  If  a  blackboard  contains  112  sq.  ft.  of  surface  and  is  4  ft. 
wide,  what  is  its  length? 

4.  How  high  is  a  room  18  ft.  long  and  16  ft.  wide,  if  its  four 
walls  contain  68  sq.  yd.? 

5.  The  area  of  a  rectangular  garden  is  14  sq.  rd.  8  sq.  yd.  4  sq. 
ft.  72  sq.  in.,  and  its  length  4  rd.  2  yd.     What  is  its  width? 

CARPETING,  PAVING,   ROOFING 
ILLUSTRATIVE  EXAMPLE 

187.  Utilizing  all  the  material,  how  many  yards  of  carpeting 
2  ft.  4  in.  wide,  are  required  to  cover  a  floor  21  ft.  long  and  16  ft. 
wide? 


COMPUTATION    OF    SURFACES  119 

SOLUTION 

21  sq.  ft.  (in  one  row  adjoining  the  length  of  the  room)  X  16 
(the  number  of  rows  in  the  width  of  the  room)  =  336  sq.  ft.  (total 
floor  area  to  be  carpeted).  3  sq.  ft.  (in  one  row  adjoining  the 
length  of  1  yd.  of  carpet)  X  2J  (the  number  of  such  rows  in  the 
width  of  1  yd.  of  carpet)  =  7  sq.  ft.  (area  of  1  yd.  of  carpet). 
Hence, 

336  sq.  ft.  (area  of  floor)  4-  7  sq.  ft.  (area  of  1  yd.  of  carpet) 
=  48  times  (1  yd.  of  carpet),  or  48  yd.  of  carpet. 

EXPLANATION.  As  many  yards  of  carpet  will  be  required  as  the  obtained 
area  of  1  yd.  of  the  given  carpet  (7  sq.  ft.)  is  contained  times  in  the  entire  floor 
area  to  be  carpeted  (336  sq.  ft.),  or  48  times  1  yard. 

SUMMARY,  (a)  The  floor  area  overlaid  by  one  yard  of  the  cover- 
ing material  is  the  multiplicand,  (b)  The  total  number  of  yards  of 
the  covering  material  is  the  multiplier,  (c)  The  total  floor  area  to  be 
overlaid  is  the  product. 

NOTE  1.  To  find  the  number  of  flagstones,  bricks,  etc.,  to  overlay  a  given 
surface,  divide  the  area  of  the  surface  to  be  overlaid  by  the  exposed  area  of  one 
unit  of  the  overlaying  material. 

EXAMPLES  FOR  PRACTISE 
Utilizing  all  the  material,  how  many  yards  will  carpet 

1.  A  room  26  ft.  long  and  20  ft.  wide;  carpet  2  ft.  6  in.  wide? 

2.  A  room  19  ft.  3  in.  long  and  16  ft.  wide;  carpet  2  ft.  9  in. 
wide? 

3.  A  room  27  ft.  8  in.  long  and  23  ft.  7  in.  wide;  carpet  1  yd. 
wide? 

4.  A  room  37  ft.  4  in.  long,  25  ft.  6  in.  wide;  carpet  2  ft.  4  in. 
wide? 

5.  What  is  the  cost  of  laying  matting  2  yd.  wide  on  the 
floor  of  a  room  24  ft.  long  and  18  ft.  wide  at  45^  per  yard? 

6.  How  many  bricks  8  in.  long,  4  in.  wide,  and  2  in.  thick,  are 
necessary  to  cover  a  pavement  18  ft.  long  and  12  ft.  wide,  if  the 
bricks  are  laid  flatwise? 

7.  Laid  edgewise,  how  many  bricks  are  necessary  to  pave  a 
stable  entrance  14  ft.  long  and  12  ft.  wide,  if  the  bricks  are  8  in. 
long,  4  in.  wide,  and  2  in.  thick? 


120  COMPUTATION    OF    SURFACES 

8.  If  placed  endwise,  how  many  bricks  are  needed  to  cover 
a  sidewalk  32  ft.  6  in.  long  and  18  ft.  4  in.  wide,  if  the  bricks  are 
8  in.  long,  4  in.  wide,  and  2  in.  thick? 

9.  How  many  granite  blocks  8  in.  long  and  8  in.  wide  are 
needed  to  cover  a  yard  18  ft.  long  and  15  ft.  wide? 

10.  How  many  pieces  of  turf   1  ft.  long  and  9  in.  wide  are 
needed  to  sod  a  plat  of  ground  62  ft.   3  in.  long  and  43  ft. 
wide? 

11.  How  many  yards  of  carpet  2  ft.  6  in.  wide  are  needed  to 
cover  the  floor  of  a  room  27  ft.  long  and  20  ft.  wide,  if  the  strips 
are  laid  crosswise  to  the  room? 

SOLUTION  EXPLANATION.     As  the  strips 

.  are  to    be  laid    crosswise    (across 

27  ft.  ^  2?  ft.  =    LO^r  times  1  strip     the  iengthof  theroom),  divide  the 

or  practically  11  strips.  len&th  of  the  room  <27  ft->  bv  the 

width  of  the  carpet  (2£  ft.)  to  find 

20  ft.  X  11  =  220  ft.  =  73J  yd.  the    number    of    required     strips 

(practically  11  strips,  the  excess 

of  the  llth  strip,  6  in.,  being  either  cut  off  before  laying  or  turned 
under  while  laying).  As  the  length  of  each  strip  must  be  the  same  as 
the  width  of  the  room,  multiply  the  length  of  each  strip  or  multiplicand 
(20  ft.)  by  the  number  of  strips  or  multiplier  (11)  to  find  the  total  number 
of  yards  of  carpet  needed  (220  ft.,  or  73 £  yd.). 

If  there  is  no  loss  in  matching  figures,  how  many  yards  of 
carpet  must  be  purchased  to  cover  a  floor 

12.  26  ft.  3  in.  long  and  20  ft.  5  in.  wide;  carpet  1  yd.  wide, 
laid  crosswise? 

13.  21  ft.  9  in.  long  and  18  ft.  8  in.  wide;   carpet  2  ft.  4  in. 
wide,  laid  lengthwise? 

14.  18  ft.  6  in.  long  and  16  ft.  9  in.  wide;   carpet  2  ft.  6  in. 
wide,  laid  crosswise? 

NOTE  2.  In  matching  figures,  an  allowance  should  be  made  for  the 
wastage  caused  by  adding  to  the  length  of  each  following  strip  as  much  length 
of  carpet  as  must  be  cut  off  to  enable  it  to  match  the  adjoining  edge  of  the  first 
strip.  Thus,  in  Ex.  11,  if  18  in.  had  been  allowed  for  wastage  in  matching 
figures,  the  first  strip  would  have  been  20  ft.  long  and  the  10  remaining  strips 
20  ft.  +  18  in.  long;  and  the  length  of  carpet  needed  would  have  been  20  ft. 
+  (2H  ft.  X  10),  or  235  ft.,  or  78 1  yd. 


COMPUTATION    OF    LUMBER  121 

Allowing  10  inches  per  strip  for  matching  figures,  how  many 
yards  of  carpet  will  be  needed  to  cover  a  floor 

15.  30  ft.  6  in.  long  and  26  ft.  3  in.  wide;  carpet  1  yd.  wide, 
laid  lengthwise? 

16.  25  ft.  3  in.  long  and  18  ft.  6  in.  wide;   carpet  2  ft.  4  in. 
wide,  laid  crosswise? 

COMPUTATION    OF    LUMBER 

188.  Lumber  measure  is  used  in  measuring  timber  after  it  has 
been  sawed  into  boards  or  planks,  scantling,  joists,  and  other 
building  material. 

189.  The  board   foot   is  the   standard  by  which   lumber  is 
measured.     A  board  foot  is  1  ft.  long,  1  ft.  wide  and  1  in.  thick; 
or  1  sq.  ft.  of  the  face  of  a  board  which  is  1  in.  thick. 

Thus,  a  board  16  ft.  long,  2  ft.  wide,  and  1  in.  thick,  is  estimated  to  con- 
tain 16  X  2,  or  32  board  feet. 

NOTE  1.  If  a  board  is  more  than  1  in.  thick,  the  number  of  square  feet 
in  its  face  should  be  proportionately  increased  in  estimating  the  number  of 
board  feet  it  contains.  Thus,  the  number  of  board  feet  in  a  plank  12  ft.  long, 
2  ft.  wide,  and  U  in.  thick  =  1 J  times  12  X  2,  or  30  bd.  ft.;  if  U  in.  thick  = 
li  times  12  X  2,  or  36  bd.  ft.;  if  2  in.  thick  =  2  times  12  X  2,  or  48  bd.  ft. 

NOTE  2.  Hewn  timber  is  sold  by  board  measure,  or  by  cubic  measure, 
and  sometimes  by  the  "running  foot."  Even  when  sold  by  the  running  foot, 
its  width  and  thickness  are  the  governing  factors  in  determining  the  price  per 
running  foot. 

NOTE  3.  The  average  width  of  a  tapering  board  is  ascertained  by  meas- 
uring midway  between  its  two  ends,  or  by  taking  hah*  the  sum  of  the  widths 
of  its  two  ends. 

NOTE  4.  In  measuring  two  or  more  boards  of  uniform  length  and  thick- 
ness, it  is  more  convenient  to  multiply  in  one  operation  the  uniform  length  by 
the  total  width  of  all  the  boards  to  obtain  the  total  board  feet. 

ILLUSTRATIVE  EXAMPLE 

How  many  bd.  ft.  in  a  plank  18  ft.  long,  6  in.  wide,  and  1?  in. 
thick? 

SOLUTION 

18  bd.  ft.  (if  1  ft.  wide  and  1  in.  thick)  X  \  (of  1  ft.  wide)  =  9 
bd.  ft.  9  bd.  ft.  (if  1  in.  thick)  X  li  (actual  thickness  in  inches) 
=  13J  bd.  ft. 


122  COMPUTATION    OF    VOLUMES 

EXPLANATION.  A  plank  18  ft.  long,  1  ft.  wide,  and  1  in.  thick  contains 
18  bd.  ft.;  and  if  only  i  of  a  foot  (6  in.)  wide,  contains  1  of  18  bd.  ft.,  or  9  bd. 
ft.;  and  if  li  in.  thick,  contains  U  times  9  bd.  ft.,  or  131  bd.  ft. 

NOTE.  5.  It  is  sometimes  more  convenient  to  multiply  the  length  in  feet 
by  the  width  in  inches  (or  12  times  the  multiplier)  to  find  the  number  of  board 
feet  in  a  1-in.  plank;  and  in  correction  to  divide  the  resulting  product  by  12. 

EXAMPLES  FOR  PRACTISE 

Find  the  number  of  board  feet  in  a  plank  which  is 


Long 

Wide 

Thick 

Long 

Wide 

Thick 

1.    16ft. 

9  in. 

1  in. 

5.   18  ft.  6  in. 

6  in. 

2  in. 

2.    18ft. 

7  in. 

1  in. 

6.    15ft.  9  in. 

7  in. 

2}  in. 

3.   14ft. 

Gin. 

IJin. 

7.   22ft. 

5  in. 

If  in. 

4.   20ft. 

8  in. 

If  in. 

8.   24ft. 

1  ft.  3  in. 

2  in. 

9.  How  many  board  feet  in  5  planks,  each  16  ft.  long  and  1  \ 
in.  thick,  if  their  respective  widths  are  6  in.,  7  in.,  9  in.,  10  in.,  and 
11  in.?  [Note  4] 

10.  How  many  board  feet  in  a  plank   18  ft.  long,   If   in. 
thick,  and  8  in.  wide  at  one  end  and  6  in.  wide  at  the  other? 
[Note  3] 

11.  What  is  the  cost  of  an  invoice  of  2-in.  planks   18  ft. 
long,  at  $24.75  per  thousand  feet,  if  their  united  width  is  45  ft. 
7  in.? 

12.  What  is  the  cost  of  24  beams  each  26  ft.  long,  10  in.  wide, 
and  3  in.  thick,  at  $21.50  per  thousand  feet? 

COMPUTATION    OF   VOLUMES 

190.  By  volume  is  meant  the  contents  included  within  the 
bounding  surfaces  of  a  given  amount  of  space. 

NOTE  1.  A  rectangular  volume  is  one  which  is  bounded  by  six  rectangular 
surfaces  [182].  In  addition  to  the  length  and  breadth  of  rectangular  surfaces, 
rectangular  volumes  have  a  third  dimension,  thickness. 

NOTE  2.  A  cube  is  a  rectangular  volume  bounded  by  six  equal  squares, 
called  faces.  All  the  edges  of  a  cube  are  of  equal  length.  Thus,  a  cubic  inch 
is  a  rectangular  volume  each  edge  of  which  is  one  linear  inch,  and  each  of  the 
six  faces  of  which  is  one  square  inch;  and  a  cubic  foot  is  a  rectangular  volume 
each  edge  of  which  is  one  linear  foot,  and  each  face  of  which  is  one  square  foot. 


COMPUTATION    OF    VOLUMES 


123 


191.  The  contents  or  volume  of  a  rectangular  solid  is  the 
amount  of  space  included  within  its  six  faces,  expressed  in  cubic 
units  of  any  convenient  denomination. 

Thus,  to  find  the  contents  of  a  rectangular  body  which  is  5  ft.  long,  4  ft. 
wide,  and  3  ft.  deep,  take  any  convenient  measuring  unit,  say,  one  cubic  foot 
(Fig.  1)  as  the  standard.  The  contents  can  then  be  found  as  follows: 

FIG.  3       I  FT. 


I  FT. 


FIG. 4 


20  CU.  FT.  X  3  =  60  CU.  FT. 

Fig.  1  is  the  measuring  unit,  which  is  1  ft.  long,  1  ft.  wide,  and  1  ft.  deep, 
or  1  cu.  ft. 

Fig.  2  shows  the  number  of  these  cubic  measuring  units  in  one  row  of  the 
upper  layer  of  Fig.  4  (4  cu.  ft.). 

Fig.  3  shows  the  number  of  these  cubic  measuring  units  in  all  the  five 
rows  of  the  upper  layer  of  Fig.  4  (4  cu.  ft.  X  5  =  20  cu.  ft.). 

Fig.  4  shows  the  number  of  these  cubic  measuring  units  in  all  three  of  the 
layers  of  the  given  body  (20  cu.  ft.  X  3  =  60  cu.  ft.). 

Hence,  a  rectangular  body  5  ft.  long,  4  ft.  wide,  and  3  ft.  deep,  contains 
5  cu.  ft.  X  4  X  3,  or  60  cu.  ft. 

SUMMARY,  (a)  The  number  of  cubic  measuring  units  in  one 
layer  (adjoining  one  face  of  a  rectangular  body)  is  the  multipli- 
cand, (b)  The  number  of  such  layers  is  the  multiplier,  (c)  The 
number  of  the  cubic  measuring  units  in  all  the  layers  (the  contents 
of  the  rectangular  body)  is  the  product. 

NOTE  1.  Before  finding  the  contents  of  a  rectangular  body,  its  length, 
breadth,  and  thickness  should  be  reduced  to  the  same  linear  denomination,  if 
not  already  so. 


124  COMPUTATION    OF    VOLUMES 

ILLUSTRATIVE  EXAMPLE 

How  many  cubic  feet  of  earth  are  in  a  rectangular  embankment 
16  ft.  6  in.  long,  10  ft.  9  in.  wide,  and  6  ft.  high? 

SOLUTION 

16£  cu.  ft.  X  10f  X  6  =  1064i  cu.  ft. 

EXPLANATION.  The  number  of  cu.  ft.  in  the  row  adjoining  the  long  edge 
of  the  lower  section  of  the  embankment  (161  cu.  ft.)  multiplied  by  the  number 
of  such  rows  in  the  lower  section  (lOf )  will  produce  the  number  of  cu.  ft.  in  all 
the  rows  of  the  lower  section  (177f  cu.  ft.);  and  the  number  of  cu.  ft.  in  all  the 
rows  of  the  lower  section  of  1  ft.  in  height  (177f )  multiplied  by  the  number  of 
such  sections  in  6  ft.  of  height  (6)  will  produce  the  number  of  cu.  ft.  in  all  the 
sections,  or  in  the  entire  embankment  (1064  \  cu.  ft.). 

EXAMPLES  FOR  PRACTISE 

1.  How  many  cubic  feet  of  wood  are  in  a  pile  28  ft.  long,  6  ft. 
high,  and  4  ft.  wide? 

2.  How  many  cubic  yards  of  air  are  in  a  room  16  ft.  long,  12  ft. 
wide,  and  10  ft.  high? 

3.  How  many  cubic  feet  of  wood  are  in  a  squared  log  18  ft. 
long,  9  in.  wide,  and  9  in.  thick? 

4.  How  many  cubic  yards  of  stone  are  in  a  rectangular  pile 
which  is  20  ft.  long,  10  ft.  wide,  and  5  ft.  high? 

5.  What  is  the  cost  of  digging  a  rectangular  excavation  24  ft. 
long,  18  ft.  wide,  and  12  ft.  deep,  at  45 i  per  cu.  yd.? 

6.  What  is  the  cost  of  a  pile  of  cord  wood  which  is  35  ft.  long, 
6  ft.  high,  and  3  ft.  wide  at  $7.15  per  cord? 

7.  A  cellar  is  18  ft.  long  and  16  ft.  wide,  and  contains  96  cu. 
yd.     What  is  its  depth? 

SOLUTION 

18  cu.  ft.  X  16  =  288  cu.  ft.      [Summary,  a.] 
27  cu.  ft.  X  96  =  2592  cu.  ft.      [Summary,  c] 
2592  cu.  ft.  ^  288  cu.  ft.  =  9  times  (1  ft.  deep). 

EXPLANATION.  A  cellar  which  is  18  ft.  long,  16  ft.  wide,  and  1  ft.  deep, 
contains  18  cu.  ft.  X  16  X  1,  or  288  cu.  ft.  Therefore  a  cellar  18  ft.  long  and 
16  ft.  wide  to  contain  96  cu.  yd.  or  2592  cu.  ft.  must  be  as  many  times  1  ft.  deep 
as  288  cu.  ft.  are  contained  times  in  2592  cu.  ft.,  that  is,  9  times  1  ft.  deep. 


COMPUTATION    OF    CAPACITIES  125 

NOTE.  2.  If  necessary,  reduce  the  given  dimensions  to  linear  units,  and 
the  given  volume  to  cubic  units,  of  the  same  denomination,  before  finding 
the  required  dimension  [Prin.  1,  34]. 

8.  A  room  48  ft.  long  and  32  ft.  wide  contains  23040  cu.  ft. 
What  is  its  height? 

9.  A  contractor  dug  a  cellar  36  ft.  long  and  12  ft.  deep,  and 
removed  336  cu.  yd.  of  earth.     What  was  its  width? 

10.  A  purchaser  paid  $59.50  for  a  pile  of  cord  wood  28  ft.  long 
and  4  ft.  wide,  bought  at  $8.50  per  cord.     How  high  was  the  pile? 

11.  A  wagon-body  12  ft.  long  and  4  ft.  4  in.  wide  contains  78 
cu.  ft.     What  is  its  height? 

12.  How  long  must  a  rectangular  embankment  be  to  contain 
35  cu.  yd.  5  cu.  ft.,  if  it  is  6  ft.  3  in.  wide,  and  9  ft.  6  in.  high? 

COMPUTATION  OF  CAPACITIES 

192.  By  capacity  is  meant  the  amount  of  space  within  a  con- 
taining vessel  which  enables  it  to  hold  a  given  quantity  of  water, 
grain,  fruit,  vegetables,  coal,  etc. 

NOTE  1.  In  finding  the  amount  of  any  given  quantity  of  grains,  seeds,  or 
berries,  stricken  measure  is  used.  By  stricken  measure  is  meant  the  measuring 
vessel  even  full  and  stricken  off  with  a  rule  or  striker.  A  stricken  bushel  con- 
tains 2150.42  cu.  in. 

NOTE  2.  In  finding  the  amount  of  any  given  quantity  of  coal,  lime,  un- 
shelled  corn,  vegetables,  or  the  larger  fruits,  heaped  measure  is  used.  Heaped 
measure  means  the  measuring  vessel  heaped  in  the  form  of  a  cone.  A  heaped 
bushel  contains  2747.71  cu.  in.  4  heaped  bushels  are  usually  considered 
equivalent  to  5  stricken  bushels. 

NOTE  3.  A  gallon,  liquid  measure,  contains  231  cu.  in.;  and  a  gallon,  dry 
measure  (ipk.),  268.8  cu.  in. 

EXAMPLES  FOR  PRACTISE 

1.  A  rectangular  bin  is  6  ft.  long,  5  ft.  wide,  and  4  ft.  deep. 
How  many  bushels  of  grain  will  it  hold? 

SOLUTION 

6  cu.  ft.  X  5  X  4  =  120  cu.  ft. 
1728  cu.  in.  X  120  =  207360  cu.  in. 
207360  cu.  in.  ^  2150.42  cu.  in.  =  96.42  +  times  1  bushel. 

EXPLANATION.  As  grain  is  measured  by  stricken  bushels  [Note  1]  divide 
the  capacity  of  the  bin  (207360  cu.  in.)  by  the  contents  of  1  stricken  bushel 
(2150.42  cu.  in.)  to  find  how  many  times  1  bu.  of  grain  the  bin  will  hold. 


126  COMPUTATION    OF    CAPACITIES 

2.  A  rectangular  bin  is  8  ft.  long,  6  ft.  wide,  and  3  ft.  9  in. 
deep.     How  many  bushels  of  potatoes  can  it  hold?     [Note  2]. 

3.  How  many  bushels  of  oats  can  be  stored  in  a  rectangular 
bin  10  ft.  3  in.  long,  4  ft.  wide,  and  3  ft.  6  in.  deep? 

4.  A  cart-body  is  7  ft.  long,  4  ft.  3  in.  wide,  and  2  ft.  9  in.  deep. 
How  many  bushels  of  shelled  corn  can  it  hold? 

5.  A  wagon-body  is  10  ft.  long,  4  ft.  2  in.  wide,  and  1  ft.  8  in. 
deep.     How  many  bushels  of  apples  can  it  hold? 

6.  How  many  gallons  of  water  can  a  rectangular  cistern  hold 
if  it  is  6  ft.  9  in.  long,  4  ft.  6  in.  wide,  and  10  ft.  deep. 

SUGGESTION.  Capacity  of  cistern  in  cu.  in.  -r-  capacity  of  1  liquid 
gallon  in  cu.  in.  [Note  3]  =  capacity  of  cistern  in  liquid  gallons. 

7.  How  many  gallons  of  water  will  fill  a  rectangular  cistern 
12  ft.  long,  12  ft.  wide,  and  4  ft.  3  in.  deep? 

8.  How  many  gallons  of  water  can  be  placed  in  a  tank  which 
is  capable  of  holding  35  bushels  of  rye? 

9.  How  many  bushels  of  potatoes  can  be  stored  in  a  vessel 
capable  of  holding  265  gallons  of  water? 

10.  A  cellar  30  ft.  long,  18  ft.  wide,  and  measuring  8  ft.  from 
the  street  level  to  the  bottom  of  the  cellar,  was  filled  by  an  over- 
flow from  a  neighboring  stream.     How  many  gallons  of  water  did 
it  contain? 

11.  What  must  be  the  depth  of  a  bin  8  ft.  long  and  6  ft.  wide, 
to  hold  96  bushels  of  oats? 

SUGGESTION.  Capacity  of  1  bu.  (2150.42  cu.  in.)  X  number  of  bu. 
(96)  =  capacity  of  the  bin.  Then  find  required  dimension  as  in  solution 
of  Ex.  7, 191. 

12.  What  must  be  the  width  of  a  wagon-body  10  ft.  long  and 
2  ft.  deep,  to  hold  53  bu.  of  turnips? 

13.  What  must  be  the  depth  of  a  rectangular  cistern  6  ft.  4  in. 
long  and  5  ft.  2  in.  wide,  to  hold  2500  gallons  of  water? 

NOTE  4.     A  sufficiently  close  approximation  to  meet  ordinary  require- 
ments may  be  obtained  by  multiplying  the  capacity  of  the  containing  vessel 
/  in  cu.  ft.  by  .8  to  find  its  capacity  in  stricken  bushels,  or  by  .63  to  find  its 
capacity  in  heaped  bushels. 

14.  A  farmer  has  a  rectangular  bin  10  ft.  long,  4  ft.  wide,  and 


COMPUTATION    OF   CAPACITIES  127 

4  ft.  deep.     What  is  the  approximate  number  of  bushels  of  wheat 
which  it  can  hold? 

15.  A  man  had  a  rectangular  bin  built  in  his  cellar  to  hold 
coal.     If  the  bin  was  10  ft.  long,  6  ft.  wide,  and  4  ft.  deep,  what 
approximate  weight  of  coal  will  it  hold,  allowing  80  pounds  to  a 
bushel  and  2240  pounds  to  a  ton? 

16.  Allowing  28  bu.  to  a  long  ton  of  2240  lb.,  how  long  must 
be  a  rectangular  coal  bin  16  ft.  wide  and  4  ft.  deep  to  hold,  approxi- 
mately, 6  tons  of  coal? 

17.  A  rectangular  bin  is  12  ft.  long,  8  ft.  wide,  and  4  ft.  deep. 
What  approximate  number  of  bushels  of  potatoes  can  it  hold? 

NOTE  5.  If  a  crib  or  bin  has  uniformly  sloping  sides,  its  average  width 
is  one-half  the  sum  of  its  top  and  bottom  widths  [Note  3,  189].  Thus,  the 
contents  of  a  granary  18  ft.  long,  12  ft.  high,  10  ft.  wide  at  the  bottom  and  14 
ft.  wide  at  the  top,  are  18  cu.  ft.  X  12  X  |  of  (10  +  14),  or  2592  cu.  ft. 

18.  How  many  bushels  of  shelled  corn  can  be  stored  in  a  crib 
which  is  12  ft.  long,  8  ft.  deep,  6  ft.  wide  at  the  bottom  and  10  ft. 
wide  at  the  top,  using  the  standard  of  Note  1? 

19.  How  many  bushels  of  turnips  can  be  stored  in  a  bin  8  ft. 
long,  6  ft.  deep,  4  ft.  wide  at  the  bottom  and  6  ft.  wide  at  the  top, 
approximating  by  Note  4? 

NOTE  6.  A  containing  vessel  is  said  to  be  cylindrical  if 
it  is  circular,  is  of  uniform  diameter  throughout  its  depth, 
and  its  ends  are  parallel.  The  capacity  of  a  cylindrical 
receptacle  is  .7854  of  that  of  a  rectangular  vessel  of  the 
same  depth,  and  having  each  of  its  two  other  dimensions 
A  CYLINDER  equal  to  the  diameter  of  the  cylindrical  receptacle. 

20.  A  cylindrical  cistern  is  12  ft.  deep  and  4  ft.  in    diameter. 
How  many  gallons  of  water  can  it  hold? 

SOLUTION 

12  cu.  ft.  X  4  X  4  =  192  cu.  ft.,  capacity  of  a  rectangular  cistern  of 
equal  depth  (12  ft.)  and  a  length  (4  ft.)  and  a  width  (4  ft.)  equal  to  the 
diameter  of  the  cylindrical  cistern.  Hence,  192  cu.  ft.  X  .7854  =  150.7968 
cu.  ft.,  capacity  of  the  given  cylindrical  cistern. 

(1728  cu.  in.  X  150.7968)  -=-  231  cu.  in.  =  1128.0384  times  1  gal.  [Note  3]. 

21.  A  cylindrical  cistern  is  6  ft.  in  diameter  and  14  ft.  in  depth. 
How  many  gallons  of  water  is  it  capable  of  holding? 


128  COMPUTATION    OF    CAPACITIES 

22.  A  well  is  4  ft.  in  diameter  and  the  water  in  it  is  6  ft. 
deep.     How  many  gallons  of  water  are  in  the  well? 

23.  A  cylindrical  bin  which  is  7  ft.  high  and  3  ft.  6  in.  in 
diameter  is  filled  with  wheat.     How  many  bushels  does  it  contain? 

24.  A  circular  railroad  tank  is  8  ft.  in  diameter  and  contains 
3500  gallons  of  water.     How  deep  is  the  water? 

NOTE  7.  A  receptacle  whose  sides  are  uniformly  round  but  tapering, 
and  whose  ends  are  parallel  but  unequal  circles,  is  equivalent  to  a  cylindrical 
receptacle  of  equal  depth  and  of  a  uniform  diameter  of  one-half  the  sum  of 
its  two  extreme  diameters.  Thus,  such  a  receptacle  if  8  ft.  in  depth,  6  ft. 
in  diameter  at  one  extreme  and  4  ft.  in  diameter  at  the  other  extreme,  is  equal 
to  a  cylindrical  vessel  which  is  8  ft.  in  depth  and  has  a  uniform  diameter 
of  (6  +  4)  -r-  2,  or  5  ft. 

25.  How  many  cubic  feet  in  a  log  20  ft.   long,   18  in.  in 
diameter  at  one  end,  and  12  in.  in  diameter  at  the  other  end? 

26.  How  many  cu.  in.  in  a  bucket  which  is  2  ft.  in  depth, 
12  in.  in  diameter  at  the  bottom,  and  16  in.  in  diameter  at  the 
top? 

27.  How  many  gallons  of  water  will  a  cistern  hold  which  is 
18  ft.  in  depth,  10  ft.  in  diameter  at  the  bottom,  and  6  ft.  in 
diameter  at  the  top? 

NOTE  8.  Owing  to  the  lack  of  uniformity  in  the  bulge  of  a  barrel  or  cask, 
its  average  diameter  is  approximately  found  by  increasing  its  head  diameter 
by  %  of  the  difference  between  its  bung  diameter  and  its  head  diameter  if  its 
staves  are  much  curved,  or  by  3  of  such  difference  if  its  staves  are  slightly 
curved.  Thus,  the  capacity  of  a  barrel  whose  length  is  36  in.,  bung  diam- 
eter 30  in.,  and  head  diameter  24  in.  will  approximately  equal  that  of  a 
cylinder  36  in.  long  and  a  uniform  diameter  of  24  in.  +  f  of  (30  in.  —  24  in.), 
or  28  in. 

28.  How  many  gallons  of  cider  will  a  barrel  hold  which  is 
40  in.  long,  bung  diameter  32  in.,  and  head  diameter  27  in? 

29.  What  is  the  capacity  in  gallons  of  a  cask  3  ft.  4  in.  long, 
head  diameter  24  in.,  bung  diameter  30  in.  ? 

30.  What  is  the  capacity  in  gallons  of  a  barrel  36  in.  long, 
bung  diameter  28  in.,  head  diameter  21  in.  ? 


PERCENTAGE 

193.  INTRODUCTION.   All  possible  arithmetical  terms  [Note,  5] 
in  all  their  various  relations  to  each  other,  and  the  means  for  com- 
pleting those   relations,  have  now  been   considered   [48-53  for 
integers,  and  107-109  for  fractions].     The  successive  subjects  from 
109  to  the  end  of  this  volume  are  simply  applications  of  these 
comprehensive  relationships  to  new  topics  which  are  of  sufficient 
importance  to  justify  special  treatment. 

In  109,  the  comparison  of  quantities  was  made  without  refer- 
ence to  any  particular  numerical  form  in  which  the  result  of  the 
comparison  was  to  be  expressed.  In  percentage,  however,  the 
result  of  the  comparison  is  limited  in  expression  to  a  particular 
form,  that  of  hundredths.  The  title  "  percentage  "  suggests  this  limi- 
tation, being  derived  from  per,  by;  centum,  hundred;  age,  that 
which.  Hence,  percentage  is  the  name  of  that  specific  class  of 
calculations,  whether  in  addition,  subtraction,  multiplication,  or 
division,  which  is  by  hundredths.  Thus,  to  add  .03  to  .06,  to  sub- 
tract .03  from  .06,  to  multiply  6  by  .03,  or  to  divide  6  by  .03, 
may  properly  be  called  operations  in  percentage.  The  name 
percentage,  however,  is  ordinarily  restricted  to  problems  which 
involve  comparisons  of  quantities  the  results  of  which  are  expressed 
in  hundredths,  as  in  the  following :  $5  are  how  many  hundredths 
of  $25?  20  hundredths  of  how  many  dollars  equal  $5?  20  hun- 
dredths of  $25  equal  how  many  dollars? 

SUGGESTION.  The  learner  is  referred  to  a,  6  and  c  of  INTRODUCTION,  109, 
for  a  full  exposition  of  the  general  treatment  of  all  numerical  comparisons, 
which  will  serve  as  a  suitable  preparation  for  considering  the  special 
application  of  the  same  general  principles  to  particular  comparisons  by 

hundredths. 

194.  Operations  in  percentage  embrace  all  calculations  which 
involve  a  specific  comparison  by  hundredths. 

195.  The  base  is  the  quantity  taken  as  the  standard  with 
which  some  other  quantity  is  to  be  compared. 

NOTE.  Any  required  comparison  of  a  quantity  must  include  a  definite 
basis  or  standard,  expressed  or  understood,  with  which  the  comparison  is  to 
be  made.  Thus,  a  given  quantity,  say  $5,  may  be  equal  to,  or  more  than,  or 


130  PERCENTAGE 

less  than,  according  to  the  standard  with  which  it  may  be  compared.  That  is, 
$5  will  be  equal  to,  if  compared  with  $5;  it  will  be  more  than,  if  compared  with 
$4;  or  it  will  be  less  than,  if  compared  with  $6. 

196.  The  rate  per  cent,  denotes  what  comparative  part  of  the 
base,  expressed  in  hundredths,  is  the  quantity  which  is  contrasted 
with  the  base. 

NOTE  1.  The  term  rate  when  not  followed  by  the  phrase  per  cent,  or  by 
the  symbol  (%)  which  means  "per  cent."  may  signify  any  part  of  the  standard 
or  base.  Thus,  in  the  expressions  4  times  $600,  f  of  $600,  .005  of  $600,  and 
.08  of  $600,  4,  f,  .005,  and  .08  are  rates;  the  first  (4)  being  an  integral  rate  per 
unit,  the  second  (f)  a  common  fractional  rate  per  third,  the  third  (.005)  a 
decimal  rate  per  thousandth,  and  only  the  fourth  (.08)  a  rate  per  hundredth  or 
per  cent. 

NOTE  2.  To  make  a  comparison  of  one  quantity  with  another  taken  as 
the  standard,  some  graduated  measuring  unit  is  necessary.  In  percentage, 
the  accepted  measuring  unit  is  1,  which  represents  the  entire  standard  of  com- 
parison and  which  for  convenience  is  conceived  to  be  graduated  or  scaled  in 
hundredths.  Thus,  in  comparing  $2  with  $5,  the  measuring  standard  or  base  is 
$5,  represented  by  the  measuring  unit  1  (meaning  1  time  $5,  or  the  entire  $5) ; 
and  this  measuring  unit  1  is  conceived  to  be  divided  or  scaled  into  100  equal 
parts.  If  $2  are  compared  to  the  standard  ($5)  by  means  of  this  graduated 
scale,  it  will  be  found  to  measure  up  to  the  fortieth  of  its  one  hundred  gradua- 
tions and  therefore  $2  are  said  to  be  40  hundredths  or  40  %  of  the  standard  ($5). 

197.  The  percentage  is  the  quantity  which  is  compared  with 
the  base. 

NOTE.  A  percentage  expresses  the  value  of  a  given  or  required  number 
of  hundredths  of  a  certain  quantity  taken  as  the  base.  Thus,  in  the  con- 
summated comparison  "$200  are  25  %  of  $800,"  $800  is  the  indicated  base  or 
standard  of  comparison;  $200  is  the  percentage  or  quantity  compared  with 
the  standard;  and  25  %  is  the  rate  %  or  result  of  the  comparison,  denoting 
what  part  $200  are  of  $800,  when  expressed  in  hundredths. 

198.  Identification  of  terms.     In  certain  classes  of  problems, 
particularly  in  "  applied  percentage,"  there  is  a  uniformly  accepted 
base,  and  it  can  therefore  be  "understood."     In  an  important 
class  of  percentage  problems,  however,  there  is  no  uniformly  rec- 
ognized standard,  and  any  base  may  be  arbitrarily  assumed,  as 
the  author  of  the  problem  may  direct.     All  bases  which  are  not 
understood  must  necessarily  be  indicated,     (a)  The  conventional 
form  for  indicating  arbitrary  bases  is  the  expression  "of "  (mean- 


PERCENTAGE  131 

ing  times)  placed  immediately  after  the  rate  %  and  immediately 
before  the  base.  (6)  It  is  sometimes  necessary  to  indicate  the 
base  with  the  purpose  of  having  it  increased  [109,  b,]  or  decreased 
[109,  c]  by  a  given  or  required  percentage,  in  which  case  the 
phrases  "more  than"  (implying  increase)  or  "less  than"  (imply- 
ing decrease),  are  substituted  immediately  after  the  rate  %  and 
before  the  indicated  base,  (c)  After  the  base  has  been  identified, 
any  quantity  which  is  placed  in  comparison  with  the  indicated 
base  with  the  view  of  ascertaining  how  many  hundredths  of  the 
given  base  it  is,  or  of  what  required  base  it  is  a  given  number  of 
hundredths,  is  called  a  percentage.  Thus,  in  the  consummated 
comparisons,  "$20  are  25%  of  $80,"  "$20  are  75%  less  than 
$80,"  the  indicated  base  is  $80,  and  the  percentage  $20;  and 
in  the  consummated  comparisons,  "$30  are  120%  of  $25,"  "$30 
are  20%  more  than  $25,"  the  indicated  base  is  $25,  and  the  per- 
centage $30. 

199.  Relation    of    terms.       (a)   The    base    is    a    multiplicand 
because  it  expresses  the  value  of  one  quantity  taken  as  the  stand- 
ard measuring  unit;    (6)  the  rate  %  is  a  multiplier  because  it  ex- 
presses the  decimal  number  (hundredths)  of   standard  measuring 
units;    (c)  the  percentage  is  a  product,  because  it  expresses  the 
value  of  all  the  considered  decimal  number  (hundredths)  of  stand- 
ard measuring  units  [Prin.  1,  2  and  3,  107]. 

Hence,  (d),  if  the  base  (multiplicand)  and  the  rate  %  (multiplier)  are 
given,  multiply  these  factors  to  find  the  percentage  (product) ;  (e)  if  the  base 
(multiplicand)  and  the  percentage  (product)  are  given,  divide  the  given 
product  by  its  given  multiplicand  factor  to  find  its  required  multiplier  factor 
or  rate;  (/)  if  the  rate  (multiplier)  and  the  percentage  (product)  are  given, 
divide  the  given  product  by  its  given  multiplier  factor  to  find  its  required 
multiplicand  factor  or  base. 

NOTE.  Though  all  rates  are  multipliers  and  all  percentages  are  products, 
learners  are  cautioned  against  dividing  any  product  by  any  multiplier  to  find 
a  particular  multiplicand.  A  product  must  be  divided  by  its  multiplier  factor, 
that  is,  by  the  particular  multiplier  which  was  used  to  produce  it,  in  order 
to  obtain  its  multiplicand  factor,  that  is,  the  particular  multiplicand  which 
was  used  to  produce  it. 

200.  Convenient  common  fractional  rates  should  be  substi- 
tuted for  inconvenient  rates  %,  particularly  for  complex  rates  % 


132  PERCENTAGE 

[117].  The  rates  %  in  common  use  which  can  be  advantageously 
changed  to  equivalent  common  fractional  rates  are  shown  in  the 
following 

TABLE  OF  EQUIVALENTS: 


A  10%  =  A  25%  =  i  62J%  =  f 

2J%  =  A-  12}%  =  |  33j%  =  i  66f%  =  | 

5%  =  A  16f%  =  }  37}%  =  |  75%  =  j 

61%  =  A  20%  =  i  50%  =  }  87}%  =  1 

NOTE.     The  tendency  to  change  indiscriminately  all  rates    %,  whether 
convenient  or  inconvenient,  to  the  form  of  common  fractions  is  unbusinesslike. 


BASE  AND  RATE  GIVEN,  TO  FIND  PERCENTAGE 

ILLUSTRATIVE  EXAMPLES 
201.   1.   Find  9%  of  $385.42. 

EXPLANATION.     $385.42  is  the  indicated   base  or  mul- 
tiplicand, being  thus  arbitrarily  shown  by  the  conventional 
$385.42       indicator   "of"   [198,  a].      9%  is  the    rate  or  multiplier, 
.09       because  it  denotes  the  decimal  number  of   times  the  base 
34.  fi87&       which  is  to-  be  expressed  by  the  required    percentage,  or 
product.      Hence,  multiply  these  two  given  factors  of  the 
required  percentage,  obtaining  $34.69—. 

2.  A  man's  salary  is  $875  per  annum.  If  his  total  expenses 
are  65  %  of  his  salary,  how  much  does  he  save? 

SOLUTION 

100  %,  rate  %  of  salary          $875,  indicated  base. 
—  65  %,  rate  %  of  expenses,   X  .35,  rate  %  saved. 
35  %,  rate  %  saved.  4375 

2625 
$306.25,  percentage  saved. 

EXPLANATION.  "His  salary"  is  arbitrarily  indicated  as  the  verbal  base 
by  being  placed  after  "  %  of,"  and  $875  is  the  numerical  expression  of  this 
verbal  base  and  therefore  the  numerical  base  or  multiplicand. 

65  %  is  the  rate  or  multiplier  of  the  base  to  produce  the  percentage  or 
product  of  total  expenses,  but  not  the  rate  or  multiplier  of  the  base  to  produce 
the  required  percentage  or  product  "saved.  If  100  %  is  the  rate  of  his  salary 


PERCENTAGE  133 

[Note  2,  196]  and  65  %  is  the  rate  of  his  total  expenses,  then  100  %  —  65  %  or 
35  %,  must  be  the  specific  rate  or  multiplier  of  the  base  ($875)  to  produce  the 
percentage  or  product  saved  ($306.25). 

3.  On  a  certain  farm  were  harvested  425  bushels  of  wheat, 
and  48%  more  oats  than  wheat.  How  many  bushels  of  oats 
were  harvested? 

EXPLANATION.     The  bushels 
SOLUTION  of  wheat   harvested  is  the  arbj. 

100  %,  rate   %  of  wheat.  trarilv  indicated  verbal  base,  be- 

,f  excess  of  oats^  -£*  "S^USS 

148%,  rate  %  of  oats  harvested.  cal  expression  of  the  bushels  of 

wheat    harvested  and  therefore 

425  bu.,  indicated  base.  the  numerical  base  or  multipli- 

1.48  rate  %  of  required  oats.  cand-     48  %   is    the   expressed 

vAr\f\  rate  or  multiplier  for  the  excess 

of  the  yield  of    oats  over  that 

of  the  wheat,  but  not  the  specific 

425  rate  or  multiplier    for  the  re- 

629700  bu.,  percentage  of  oats.  Quired    number  of    bushels   of 

oats  harvested.      Expressing  the 

indicated  base  (425  bu.)  upon  a  scale  of  hundredths  as  shown  in  Note 
2,  196  (100  %),  the  bu.  of  oats  harvested  expressed  upon  the  same  scale 
will  be  48  %  more  than  100  %,  or  148  %.  Hence,  multiply  the  indicated 
base  or  multiplicand  (425  bu.)  by  the  specific  rate  %  or  multiplier  of  the  oats 
harvested  (1.48),  obtaining  629  bu.  as  the  specific  percentage  or  product  of 
the  oats  harvested. 

RULE.  To  find  any  desired  percentage,  multiply  the  base  by 
the  specific  rate  %  of  the  desired  percentage. 

NOTE  1.  Any  given  rate  should  be  identified  not  only  generally  as  a  rate 
or  multiplier,  but  also  specifically  as  the  rate  or  multiplier  of  the  indicated 
base  to  produce  a  particular  percentage  or  product.  Thus,  in  111.  Ex.  2,  100  % 
is  the  specific  rate  of  his  salary,  65  %  is  the  specific  rate  of  his  expenses,  and  35  % 
is  the  specific  rate  of  what  is  saved.  In  111.  Ex.  3,  100  %  is  the  specific  rate  of 
the  bu.  of  wheat  harvested,  48  %  is  the  rate  of  so  much  of  the  bu.  of  oats  harvested 
as  is  more  than  the  bu.  of  wheat  harvested,  and  148  %  is  the  rate  of  the  oats 
harvested. 

NOTE  2.  Two  or  more  rates  must  be  expressed  in  terms  of  (upon  the  same 
graduated  scale  as)  the  indicated  base  before  they  can  be  added  [Prin.  1,  18] 
or  subtracted  [Prin.  1,  22]. 


134  PERCENTAGE 


EXAMPLES  FOR  PRACTISE 

1.  .07  times  $800  =  how  much?  2.  8  %  of  $300  (that  is, 
.08  times  $300)  =  what  percentage  (that  is,  what  product  by 
hundredths)?  3.  28  %  of  650  Ib.  =  how  many  pounds?  4.  84% 
of  850  men  =  how  many  men?  5.  66|  %  of  471  sheep  =  how 
many  sheep?  6.  125  %  of  584  gal.  =  how  many  gallons?  7.  J  % 
of  $6400  =  how  much?  8.  J  %  of  800  tons  =  how  many  tons? 
9.  28  %  of  £618  2s.  6d.  =  what  percentage?  10.  Find  38  %  more 
than  650  gallons.  11.  Find  25  %  more  than  716  sheep.  12.  Find 
50  %  more  than  128  horses.  13.  Find  20  %  less  than  315  days. 
14.  Find  33J  %  less  than  714  miles.  15.  Find  16|  %  less  than 
468  yards. 

16.  A  grocer  purchased  625  pounds  of  coffee  and  sold  64  %  of 
his  purchase.     How  many  pounds  of  coffee  did  he  sell? 

17.  A  ranchman  had  475  cattle,  but  lost  28  %  of  them  in  a 
blizzard.     How  many  cattle  did  he  save? 

18.  By  installing  new  machinery,  a  factory  was  enabled  to 
produce  38  %  more  than  at  first.     If  its  former  capacity  was  6500 
yards  of  cloth  per  week,  what  is  its  present  capacity? 

19.  A  man's  taxable  property  is   assessed   at   $28500,  43  % 
of  his  assessment  being  upon  his  real  estate,  and  the  remainder 
upon  his   personal   property.     What   is   the   assessed   value   of 
each? 

20.  In  a  certain  graded  school  375  pupils  are  enrolled,  28% 
of  the  pupils  being  in  the  primary  department,  and  the  remainder 
in  the  intermediate.     How  many  pupils  were  enrolled  in  the 
intermediate  department? 

21.  By  the  census  of  1910,  it  was  found  that  the  population 
of  a  certain  city  was  43  %  greater  than  by  the  census  of  1900.     If 
its  population  in  1900  was  46800,  what  was  it  in  1910? 

22.  A  farm  contains  700  acres,  25  %  of  it  being  sown  in  wheat, 
32%  being  planted  in  corn,  and  the    remainder  set  apart  for 
pasturage.      How  many  acres  of  pasture  land  does  the   farm 
contain? 

23.  A  man  died,  leaving  an  estate  worth  $128500,  54%  of 
his   estate   having   been  bequeathed   to   his  wife,    32%   to    his 


PERCENTAGE  135 

children,  and  the  remainder  to  an  asylum.  If  the  cost  of 
administering  upon  the  estate  was  5%  of  its  value,  how  much 
did  the  asylum  receive? 

BASE  AND  PERCENTAGE  GIVEN,   TO  FIND  RATE 
ILLUSTRATIVE  EXAMPLES 

202.     1.   What  %  of  $428  are  $184.04? 

EXPLANATION.    $428    is    the    arbi- 
SOLUTION  trarily  indicated  base  or  multiplicand, 

$184.04  -T-  $428  =  .43  =  43%   beinS  Placed  immediately  after  "  %  of" 

(meaning    hundred ths    times).     $184.04 

is  the  percentage  or  product,  it  being  the  quantity  which  is  placed  in 
comparison  with  the  indicated  base  to  find  what  part  of  the  base  it  is  when 
expressed  in  hundredths  [197].  If  stated  in  more  familiar  language,  the 
problem  would  read,  "How  many  hundredths  times  $428  are  $184.04." 
Hence,  divide  the  given  percentage  or  product  ($184.04)  by  its  given  base 
or  multiplicand  factor  ($428)  to  find  its  required  rate  or  multiplier  factor, 
carrying  the  division  to  hundredths  (by  so  arranging  the  percentage  that 
it  will  have  two  more  decimal  places  than  those  of  the  base),  obtaining 
.43  or  43  %. 

2.  A  has  $350  and  B.  $448.    What  %  more  than  A  has  B? 

EXPLANATION.     A's  money  is 

SOLUTION  the  indicated  verbal  base  or  mul- 

$448— $350  =  $98,  percentage  more,  tiplicand,  as  it  is  placed  imme- 
$98.00-$350=.28,or28%(more).  diatdy  after  «%  more  than" 

[198,  61,  and  $350  is  the  numerical 

expression  of  A's  money  or  the  numerical  base.  $448  is  a  percentage 
or  product,  it  being  the  quantity  which  is  placed  in  comparison  with  the 
indicated  base  to  find  what  %  more  than  the  base  it  is.  Though  B's 
money  is  a  percentage,  it  is  not  the  percentage  or  product  of  which  it  is 
required  to  find  the  rate  or  multiplier.  Therefore,  subtract  A's  money 
($350)  from  B's  money  ($448)  to  find  the  percentage  more  ($98),  or  the 
specific  percentage  or  product  of  which  it  is  required  to  find  the  specific 
rate  %  or  multiplier.  Then  divide  the  percentage  or  product  more  ($98) 
by  its  indicated  base,  or  multiplicand  factor  ($350),  to  find  its  required 
rate  or  multiplier  factor,  carrying  the  division  to  hundredths  as  in  111.  Ex.  1, 
to  obtain  the  rate  in  hundredths  (28  %). 

RULE.    Divide  the  specific  percentage  or  product  of  which  it 
is  required  to  find  the  rate,  by  its  base  or   multiplicand  factory 


136  PERCENTAGE 

and  the  quotient  carried  to  hundredths  will  be  its  required  rate  % 
or  multiplier  factor. 

NOTE  1.  Percentages  should  be  identified  so  specifically  as  to  distinguish 
each  percentage  from  all  other  percentages  by  naming  its  individual  charac- 
teristic. Thus,  in  111.  Ex.  2,  $448  and  $98  are  not  only  to  be  regarded  as  per- 
centages generally,  but  $448  as  the  specific  percentage  of  B's  money,  and  $98 
as  the  specific  percentage  more  that  B's  money  is  than  A's. 

NOTE  2.  To  obtain  a  rate  in  the  particular  form  of  a  %,  the  percentage 
should  contain  exactly  two  decimal  places  more  than  the  base.  If  the  per- 
centage contains  fewer  than  the  required  number  of  decimal  places,  supply 
the  deficiency  by  annexing  decimal  O's;  and  if  the  percentage  contains  more 
than  the  required  number  of  decimal  places,  offset  the  excess  by  annexing 
decimal  O's  to  the  base. 

NOTE  3.  To  find  the  rate  of  a  given  percentage  means  simply  to  find 
what  multiplier  of  the  indicated  base  will  produce  the  given  percentage. 

EXAMPLES  FOR  PRACTISE 

1.  $30  are  how  many  times  $6?  2.  $20  are  how  many  hun- 
dredths times  $80?  3.  $40  are  what  %  of  (that  is,  how  many 
hundredths  times)  $120?  4.  $23.29  are  what  %  of  $68.50? 
5.  $88.69  are  what  %  of  $253.40?  6.  What  %  of  $872.60  are 
$741.71?  7.  What  %  of  $5249.50  are  $3254.69?  8.  What  % 
of  $528  are  $1.32?  9.  What  %  more  than  $68.40  are  $85.50? 
10.  $322.02  are  what  %  more  than  $268.35?  11.  $275.17  are 
what  %  less  than  $786.20?  12.  What  %  less  than  $687.50  are 
$522.50? 

13.  A  grazier  owned  350  sheep,  and  wolves  destroyed  14  of 
them.     What  %  of  his  flock  did  he  lose? 

14.  A  clerk's  salary  is  $1200  per  annum.     If  he  pays  $20  per 
month  for  board  and  room  rent,  $156  per  annum  for  clothing,  and 
$300  per  annum  for  other  expenses,  what  %  of  his  salary  does  he 
save? 

15.  A  grocer  sold  1653  Ib.  sugar  and  had  522  Ib.  remaining. 
What  %  of  his  original  stock  of  sugar  did  he  sell? 

16.  A  son  inherited  920  acres  from  his  father,  and  afterwards 
sold  138  acres.  What  %  of  his  inheritance  did  he  have  remaining? 

17.  A  man  owns  real  estate  valued  at  $23223.75  and  personal 
property  valued  at  $17650.05.     What   %  less  than  the  value  of 
his  real  estate  is  the  value  of  his  personal  property? 


PERCENTAGE  137 

18.  Three  men  formed  a  copartnership,  A  investing  $11900, 
B  $14875,  and  C  $15725.     What  %  of  the  capital  of  the  firm  did 
each  invest? 

19.  The  sailing  distance  from  one  seaport  to  another  is  8750 
miles.     What  %  of  the  voyage  remained  after  a  vessel  had  sailed 
21  days  at  an  average  speed  of  150  miles  per  day? 

20.  A  merchant  engaged  in  business  with  a  capital  of  $42500 
and  at  the  end  of  the  first  year  found  that  he  had  gained  $7650. 
What  %  of  his  original  capital  did  he  gain? 

21.  A  man  borrowed  $4500  and  made  three  partial  payments 
thereon  of  $1820,  $1230,  and  $415.     What   %  of  his  original 
debt  remained  unpaid? 


PERCENTAGE  AND  RATE  GIVEN,   TO  FIND  BASE 
ILLUSTRATIVE  EXAMPLES 

203.  1.  A  man  bought  a  house  and  paid  45  %  of  the  purchase 
money  in  cash.  At  what  price  did  he  buy  the  house,  if  his  cash 
payment  was  $3825? 

SOLUTION  EXPLANATION.     The  purchase  money  is 

indicated  as  the  verbal  base  or  multiplicand 

$3825.00  -T-  .45  =  $8500      by  being  placed  immediately  after  "  %  of." 

$3825  is  a  percentage  or  product  because  it 

is  placed  in  comparison  with  the  indicated  verbal  base  (the  purchase  money). 
45  %  is  the  specific  rate  or  multiplier  for  $3825  because  it  denotes  the  result 
of  the  comparison  of  $3825  with  the  indicated  base  ($3825  being  .45  of  the 
purchase  money).  Hence,  divide  the  given  percentage  or  product  ($3825) 
by  its  specific  rate  or  multiplier  factor  (.45)  to  find  its  specific  base  or  mul- 
tiplicand factor  ($8500).  Expressed  in  more  familiar  terms,  the  question  is 
"45  hundredths  times  what  purchase  money  equal  $3825?" 

2.   A  has  $296.80,  which  is  36  %  less  money  than  B  has.     How 
much  money  has  B? 

SOLUTION 

100  %  =  rate  of  B's  money,  the  indicated  verbal  base. 
36%  =  rate  which  A's  money  is  less  than  B's. 
64  %  =  specific  rate  of  A's  money,  expressed  in  terms  of  B's. 
$296.80  (A's  money)  -r-  .64  (specific  rate  of  A's  money)  = 
$463.75  (B's  money,  or  the  base). 


138  PERCENTAGE 

EXPLANATION.  B's  money  is  indicated  as  the  required  verbal  base  by 
being  placed  immediately  after  "  %  less  than  "  [198,  6],  $296.80  is  a  percentage 
or  product  because  it  is  placed  in  comparison  with  the  indicated  base  (B's 
money) ;  and  specifically  it  is  the  percentage  of  A's  money.  36  %  is  the  specific 
rate  or  multiplier  of  what  A's  money  is  less  than  B's,  but  not  the  specific  rate  or 
multiplier  of  the  base  (B's  money)  to  produce  the  specific  percentage  or 
product  of  A's  money  ($296.80).  To  obtain  the  specific  rate  of  A's  money 
in  terms  of  B's,  let  100  %  represent  the  indicated  base,  B's  money  [Note  2, 
196];  and  as  A's  money  is  36  %  less  than  B's,  so  36  %  less  than  100  %,  or  64  %, 
is  the  specific  rate  of  A's  money.  Hence  divide  the  specific  percentage  or 
product  of  A's  money  ($296.80)  by  its  specific  rate  (the  specific  multiplier  of 
B's  money  to  produce  A's,  64  %)  to  find  the  required  base  or  multiplicand  (B's 
money,  $463.75). 

RULE.  Divide  the  given  percentage  or  product  by  its  rate  factor 
(the  multiplier  of  the  base  which  was  used  to  produce  the  given 
percentage),  and  the  quotient  mil  be  its  base  factor  (the  multipli- 
cand which  was  used  to  produce  the  given  percentage). 


EXAMPLES  FOR  PRACTISE 

1.  $63  are  7  times  how  many  dollars?  2.  72  gallons  are  .09 
times  how  many  gallons?  3.  42  yd.  are  6%  of  (.06  times)  how 
many  yards?  4.  252  tons  are  45  %  of  what  weight?  5.  28  %  of 
how  much  vinegar  =  150.64  gal.?  6.  24  %  of  what  sum  of  money 
=  $2039.10?  7.  |%  of  what  base  =  $23.82?  8.  }%  of  how 
much  money  =  $116.91? 

9.  $633.45  are  64  %  more  than  how  much  money?  10.  24  % 
more  than  what  sum  of  money  =  $928.45?  11.  52%  less  than 
what  sum  of  money  =  $616.20?  12.  £187  10s.  are  76  %  less  than 
how  much? 

13.  After  selling  414  tons  of  coal,  a  dealer  found  that  he  had 
52  %  of  his  original  stock  remaining.     How  many  tons  did  he  have 
at  first? 

14.  A  merchant  paid  $247.25,  which  was  43  %  of  his  cred- 
itor's  claim   against   him.     How   much  of  the  claim  remained 
unpaid? 

15.  A's  farm  contains  783  acres  of  land,  which  is  8%  more 
than  the  number  of  acres  in  B's  farm.     How  many  acres  does 
B's  farm  contain? 


PERCENTAGE  139 

16.  A  owns  12  %  of  the  stock  of  a  corporation,  B  15  %,  C  20  %, 
D  25  %,  and  E  the  remainder.     What  is  the  value  of  B's  stock, 
if  E's  is  worth  $7084?    \ 

17.  A's  house  is  worf^i  46%  more  than  B's  farm.     What  is 
the  value  of  B's  farm,  if  A'a  house  is  worth  $9855? 

18.  A  man  purchased  a  building  lot  for  $4800,  erected  a  house 
thereon  which  cost  $5600,  expended  $2050  in  painting  and  deco- 
rating the  house,  $510  in  furnishing  it,  and  had  28  %  of  his  money 
remaining.     How  much  did  he  have  at  first? 

19.  By  an  unfortunate  investment  a  merchant  lost  $16872, 
which  was  37  %  of  his  original  capital.     What  was  his  subsequent 
capital? 

20.  A  dealer  sold  an  invoice  of  goods  for  26%  more  than  he 
paid  for  them,  and  received  $818.37.     How  much  did  the  dealer 
pay  for  the  goods  sold? 


COMMON  APPROXIMATIONS  IN  PERCENTAGE 

204.  INTRODUCTION,  (a)  After  obtaining  a  denominate  per- 
centage by  201  or  a  denominate  base  by  203,  it  is  customary 
among  business  men  to  reduce  the  decimal  part  of  the  result  to 
the  form  of  a  common  fraction  as  in  125.  If  no  special  reason 
exists  for  reducing  the  decimal  to  a  common  fraction  of  a  partic- 
ular denominator,  it  is  usually  reduced  to  8ths.  After  thus 
reducing  the  decimal  to  8ths,  the  obtained  fraction  should  be 
expressed  in  lowest  terms,  if  not  already  so. 

(b)  In  some  lines  of  business  a  special  denominator  is  adopted 
which  expresses  the  number  of  units  of  the  next  lower  denomina- 
tion used  in  that  kind  of  business  which  equal  1  integral  unit  of 
the  obtained  result.  Thus,  coal  dealers  reduce  decimal  parts 
of  a  result  in  tons  to  the  denominator  which  expresses  the  num- 
ber of  pounds  (the  next  lower  denomination  to  tons  used  in  the 
coal  business)  which  equal  1  ton,  that  is,  to  2000ths,  if  short 
tons,  or  to  2240ths,  if  long  tons;  grain  dealers  reduce  deci- 
mal parts  of  an  obtained  result  in  bushels  to  the  denominator 
which  expresses  the  number  of  pounds  (the  next  lower  denom- 
ination to  bushels  used  in  the  grain  business)  which  equal  1 
bushel  of  that  particular  kind  of  grain,  that  is,  to  60ths  if  wheat, 
to  56ths  if  corn,  etc.  In  the  same  manner  fractional  parts  of 
other  units  of  merchandise  are  expressed  with  still  other  denom- 
inators. 


140  PERCENTAGE 

ILLUSTRATIVE  EXAMPLES 

1.  Find  8  %  of  419  bu.  wheat. 

SOLUTION 

419  EXPLANATION.     First  find  8  %  of  419  bu.  by  201, 

Qg  obtaining  33.52  bu.     Retain  the  integral  part  of  the 

result  (33  bu.)  as  a  part  of  the  final  answer,  and 

DU.  33.52  reduce  the  decimal  part  (.52  bu.)  to  pounds  (GOths) 

60  by  125,  obtaining  31  +  lb.,  thus  making  the  com- 

lb.     31  20  plete  answer  33  bu.  31  lb.,  usually  written  3331  bu. 

Ans.  33  bu.  31  lb. 

2.  A  dealer  sold  127  gal.  molasses  which  was  43  %  of  all  the 
molasses  he  had  in  stock.     How  many  gallons  did  he  have  remain- 
ing, reducing  any  decimal  of  a  gallon  to  8ths  (pints)? 

SOLUTION 

43)127.00(295|  gallons,  (base)  EXPLANATION.  Divide  the  speci- 

gg  fie  percentage  sold  (127  gal.)  by  the 

— 7  specific  rate  %  sold  (.43),  obtaining 

295  gal.  as  the  indicated  base,  or  all 

387           295 f  gal.  at  first  the  molasses  he  had  in  stock,  and  a 

230         127  gal.  sold  remainder  of  15  (if  gal.).     Multiply 

•— —                       .    .  this  remainder  (15)  by  the  required 

215        168¥  gal.  remaining  denominator  (8ths)  and  divide  the 

15  result  (120  eighths)  by  the  original 

g  divisor  (43)  obtaining  2  eighths  and 

more  than  half  of  another  eighth,  or 

43)120(3  approximately  =  practically  }  [86],  making  the  com- 

129  plete  base  295f  gal. 

EXAMPLES  FOR  PRACTISE 

1.  A  man  bought  765  bu.  potatoes  and  sold  52  %  of  his  pur- 
chase.    How  many  bushels  and  pounds    (GOths)    did  he  have 
remaining? 

2.  A  coal  dealer  bought  163  tons  of  coal  and  sold  37  %  of  his 
purchase  at  one  time,  and  45%  of  the  remainder  at  another 
time.     How  many  tons  and  pounds  (2240ths)  did  he  have  finally 
remaining? 

3.  A  grain  dealer  sold  32  %  of  his  stock  of  corn  and  had  416 
bushels  remaining.     How  many  bushels  and  pounds  (56ths)  of 
corn  did  he  sell? 


PERCENTAGE  141 

4.  A  merchant  sold  15  %  of  an  invoice  of  muslin  and  had  473 
yards  remaining.    How  many  yards  of  muslin  were  in  the  invoice, 
expressing  any  resulting  fraction  of  a  yard  in  4ths? 

5.  A  man  sold  785  bu.  of  oats,  which  was  62  %  of  his  original 
purchase.     How  many  bushels  and  pounds  (32nds)  did  he  have 
remaining? 

6.  The  distance  between  two  cities  is  268  miles.    If  a  man 
has  traveled  53  %  of  the  distance,  how  many  miles  and  yards 
(1760ths)  has  he  yet  to  travel? 

7.  A  woodman  delivered  173  cords  of  wood,  which  was  72% 
of  the  amount  specified  in  his  contract.     How  many  cords  and 
cord  feet  (8ths)  remained  to  be  delivered? 

APPROXIMATIONS  OF  RATES  % 

205.  In  the  practical  application  of  202  to  business  calcula- 
tions, many  difficulties  will  present  themselves,  methods  for  sur- 
mounting some  of  which  will  now  be  considered. 

(a)  When  results  are  obtained  in  United  States  money,  they 
are  expressed  approximately  to  the  nearest  cent.     This  will  fre- 
quently cause  embarrassment  to  the  learner.    Thus,  in  a  business 
calculation,  5%  of  $28.37,  or  $1.4185,  would  be  entered  in  the 
books  of  account  as  $1.42.     If  occasion  should  afterward  arise 
to  learn  what  %  of  $28.37  are  $1.42,  the  exact  result  (5%)  cannot 
be  obtained  by  the  usual  process.     The  insignificant  remainder 
(.0015)    when    compared    with    the    divisor 
SOLUTION  I        (28.37)   would  at  once  remind  the  experi- 
$28.37)$!. 4200(.05    enced  calculator  that  it  was  caused  by  count- 
1  4185          ing  a  fraction  of  a  cent  more  than  a  half 
15          (.85  of  a  cent)  as  a  whole  cent,  thus  making 
the  dividend  ($1.42)  .15  of  a  cent  greater 
than  the  exact  expression,  and  causing  the  remainder  of  .15  of  a 
cent  when  the  exact  quotient  (5%)  is  found. 

(6)  On  the  other  hand,  if  in  a  practical  calculation,  it  is  re- 
quired to  take  36%  of  $78.59,  the  exact  result  will  be  $28.2924, 
and  its  practical  expression  $28.29.     If  for 
any  reason,  it  is  subsequently  necessary  to 

learn  what  rate  %    had  been  emPloved  to 
obtain  this  percentage,  the  exact  result  will 
not  be  secured,  but  only  35%   with  a  very 
4  7130          significant    remainder.     This    significant    re- 
3  9295          mainder    (7835)    when   contrasted  with   the 
7835          divisor  (7859)  should  remind  the  calculator 


142  PERCENTAGE 

that  the  remainder  was  caused  by  discarding  from  the  percentage 
($28.29)  a  fraction  of  a  cent  less  than  a  half  (.24  of  a  cent),  thus 
making  the  percentage  lack  .24  of  a  cent  of  being  exactly  36%  of 
the  base,  and  causing  the  remainder  (.7835)  to  lack  .24  of  a  cent 
of  exactly  containing  the  divisor  1  more  time  than  35%,  or  36%. 
(c)  Again,  it  may  be  necessary  to  take  6^%  of  $49.32,  of 
which  the  exact  percentage  is  $3.0825,  or,  practically  expressed, 

$3.08.     If,    afterward,    it    should   be- 

SOLUTION  III  come  necessary  to  find  what  %   had 

$49  32) $3  0800 (61(7  been  used  to  obtain  $3.08,  the  result 

2  9592  would  be  6%  and  an  intermediate  re- 

—  mainder  (that  is,  neither  an  insignifi- 

^  cant  remainder  as  in  a,  nor  a  significant 

—     *  remainder   as  in  b,  but  intermediate 

9664(;  =  f  =  f     between  the  two)  of  .1208.     The  cal- 

4932  culator  should  then  at  once  conclude 

4732  that  there  was  a  fraction  of  a  per  cent 

used  in  the  rate.     The  fractional  rates 

in  common  use  are  halves,  fourths,  eighths,  and  thirds.  Of  these, 
the  last  (thirds)  will  immediately  suggest  itself  by  being  associated 
with  16,  33  and  66  in  forming  the  frequently  used  commercial  rates 
16f  %,  33J%,  and  66f  %.  If  thirds  are  not  thus  suggested  as  in 
the  present  instance,  the  remainder  (1208)  should  be  reduced  to 
eighths,  the  highest  multiple  of  the  remaining  commercial  denomi- 
nators, by  86,  obtaining  1  eighth  and  the  significant  remainder 
4732  which  lacks  200  (8  times  the  discarded  $.0025)  of  being  an- 
other eighth,  making  f ,  or  in  lowest  terms  i;  thus  showing  that  the 
rate  used  was  exactly  6J%. 

NOTE.  If  in  c,  thirds  are  suggested  by  association  with  16,  33  or  66, 
pupils  are  cautioned  against  accepting  the  probability  for  a  certainty.  The 
probability  should  be  tested,  and  only  an  insignificant  remainder  as  in  Solu- 
tion 1,  or  a  significant  remainder  as  in  Solution  2,  can  be  accepted  as  con- 
clusive. The  correct  fractional  rate  will  never  leave  an  intermediate  remainder. 

SUMMARY.  In  finding  rates  %,  insignificant  remainders  should 
be  disregarded;  significant  remainders  which  approximate  closely  to 
the  base  require  the  last  quotient  figure  to  be  increased  by  1:  and 
intermediate  remainders,  when  thirds  are  not  suggested,  should  be 
reduced  to  eighths  and  the  result  expressed  in  lowest  terms. 

NOTE.  If  a  base  expresses  many  thousands  of  dollars  as  in  great  whole- 
sale transactions,  or  millions  of  dollars  as  in  many  calculations  of  railroad 
companies  and  other  great  corporations,  rates  are  usually  carried  to  five  deci- 
mal places,  the  first  two  places  after  the  point  expressing  %,  and  the  following 
three  places  denoting  thousandths  of  a  per  cent. 


PERCENTAGE  143 

EXAMPLES   FOR  PRACTISE 

What  %  of  What  %  of 

1.  $374.86  are  $157.44?  5.  $9248  are  $3479.56? 

2.  $968.43  are  $377.69?  6.  $9.65  are  $4.04? 

3.  $758.65  are  $216.22?  7.  $67841.35  are  $42740.05? 

4.  $4832.72  are  $4687.74?  8.  $6873.24  are  $2019.01? 

9.  The  gross  earnings  of  a  railroad  for  the  year  1910  were 
$1375487.60,  and  its  operating  expenses  $296785.37.  What  %  of 
the  gross  earnings  were  the  operating  expenses  (to  5  decimal 
places)? 

10.  A  wholesale  firm  extended  credits  to  the  amount  of 
$3468756.75  during  the  year  1910,  of  which  it  failed  to  collect 
$12378.40.  What  %  of  its  credits  did^it  collect  (to  5  decimal 
places) ? 

BASES  UNDERSTOOD 

206.  INTRODUCTION.  The  idea  expressed  by  the  term  "base" 
as  employed  in  percentage  differs  little  from  that  of  the  same 
word  in  ordinary  usage.  Thus,  the  base  of  any  object,  say  a 
monument,  is  the  foundation  or  pedestal,  upon  which  it  rests,  and 
without  which  it  cannot  stand.  So  in  a  percentage  problem,  if 
the  base  is  not  directly  indicated,  it  is  " understood"  to  be  the 
entire  quantity  of  which  the  rate  %  represents  a  part,  and  of  which 
the  percentage  is  a  part,  and  therefore  that  upon  which  both 
these  terms  depend  for  their  value  as  a  monument  depends  upon 
its  pedestal  for  its  stability.  A  base  which  is  thus  comprehensible 
is  sometimes  called  a  "normal  base"  to  distinguish  it  from  an 
"arbitrary  base"  which  has  no  special  characteristic  by  which  it 
can  be  identified,  but  which  depends  simply  upon  the  arbitrary 
choice  of  the  author  of  the  problem  in  selecting  it  as  such.  A 
normal  base  is  therefore  the  quantity  from  which  the  percentage 
naturally  derives  its  origin,  that  of  which  one  would  readily  think 
as  the  whole  while  considering  any  part  of  it,  that  which  would 
naturally  suggest  itself  to  the  mind  as  the  answer  if  the  interroga- 
tive "of  what"  be  placed  immediately  after  the  rate  %. 

NOTE.  If  there  is  a  possibility  of  the  slightest  doubt  relative  to  the  identity 
of  the  base,  it  should  be  indicated  by  placing  of,  more  than,  or  less  than,  or  words 
of  equivalent  import,  immediately  after  the  rate  %  and  immediately  before 
the  otherwise  doubtful  base  [198,  a,  b,  c]. 


144  PERCENTAGE 

EXAMPLES   FOR  PRACTISE 

1.  A  man's  income  is  $1800  per  annum.     If  he  pays  20  %  for 
board,  15  %  for  clothing,  and  25  %  for  incidentals,  how  much  does 
he  save? 

2.  A  town  has  a  population  of  10500,  55  %  being  native  born, 
20  %  Germans,  15  %  Irish,  and  the  remainder  of  other  national- 
ities.    How  many  of  each  does  the  town  contain? 

3.  A  man  bought  a  house  for  $6500,  paying  $1800  in  cash, 
and  agreeing  to  settle  the  remainder  in  annual  payments  of  25  %. 
What  was  the  amount  of  each  annual  payment? 

4.  A  man  died  leaving  property  of  the  value  of   $13500. 
He  bequeathed  45%  to  his  wife,  35%  to  his  only  child,  and 
the   remainder   to   a   hospital.     How   much   did   the    hospital 
receive? 

5.  A  farm  of  105  acres  is  only  28  %  as  large  as  the  farm  adjoin- 
ing it.     How  many  acres  in  the  adjoining  farm? 

6.  If  coffee  loses  15  %  in  roasting,  how  many  pounds  of  green 
coffee  must  be  bought  to  produce  1020  pounds  when  roasted? 

7.  A  is  worth  $38500  and  B  only  47  %  as  much.     What  is  B 
worth? 

8.  A  man  weighed  190  pounds  before  an  attack  of  illness, 
and  133  pounds  directly  after  the  attack.     What  %  did  he  lose 
in  weight? 

PERCENTAGES  OF  INCREASE  OR  DECREASE 

207.  INTRODUCTION.  No  increase  or  decrease  is  possible,  except 
there  be  first,  as  a  basis  for  such  a  modification,  a  quantity  to  be 
augmented  or  diminished.  Hence,  in  all  problems  of  this  character, 
the  normal  base  always  understood  is  the  quantity  to  be  increased 
or  diminished,  that  is,  the  quantity  immediately  before  such  a 
change  has  been  effected;  and  the  normal  percentages  always 
understood  are  the  increase  or  decrease  itself,  and  the  quantity 
after  such  a  change  has  been  made.  Most  of  the  applications 
of  percentage  from  209  to  291,  in  which  bases  are  understood, 
depend  upon  this  important  principle  in  the  proper  selection  of 
the  base;  profit  being  but  another  name  for  increase,  loss  for 
decrease,  the  various  discounts  (bank,  true,  or  trade)  for  de- 
creases, etc.,  etc. 


PERCENTAGE  145 

NOTE.  Increase  means  more  than,  and  decrease  less  than  [198,  6].  Hence, 
when  in  doubt,  the  rational  answer  to  the  interrogative  "more  than  what" 
placed  after  a  rate  of  increase,  or  "less  than  what"  placed  after  a  rate  of 
decrease,  will  suggest  the  base  understood. 

EXAMPLES   FOR  PRACTISE 

1.  At  the  census  of  1900,  a  city  had  a  population  of  37240, 
and  at  the  census  of  1910,  its  population  was  50274.     What  was 
the  %  of  increase? 

2.  During  a  financial  depression,  a  clerk  had  his  salary  dimin- 
ished 15  %,  or  $300.     What  was  his  salary  after  the  reduction? 

3.  By  a  rise  in  real  estate,  a  house  has  increased  in  value  28  %, 
and  is  now  worth  $10880.      What  was  its  value  before  the  rise? 

4.  A  firm's  gross  sales  last  year  were  $287300,  and  this  year 
$333268.     What  was  the  %  of  increase? 

6.  By  introducing  new  machinery  a  factory  has  increased  its 
output  16f  %,  and  now  produces  182753  yards  of  sheeting  per 
week  more  than  before.  What  was  the  capacity  of  the  factory 
before  installing  the  new  machinery? 

6.  The  capital  of  a  firm  was  increased  20  %  in  1910,  and  15  % 
in  1911,  the  total  increase  during  the  two  years  being  $17406. 
What  was  its  capital  on  Jan.  1,  1910? 

7.  The  net  earnings  of  a  corporation  were  diminished  25% 
during  the  financial  depression  of  1907,  and  still  further  dimin- 
ished 10  %  during  the  year  1908,  but  during  the  year  1909,  under 
improved  conditions,  they  were  increased  20  %.  If  the  net  earnings 
for  1909  were  $268764.50,  what  were  the  net  earnings  during  1907? 

8.  If  the  net  capital  of  a  firm  was  $45000  on  Jan.  1,  1908, 
and  there  was  a  net  increase  of  10  %  per  annum  for  four  success- 
ive years,  what  was  its  capital  on  Jan.  1,  1912? 

REVIEW  OF  PERCENTAGE 

208.  1.  A  merchant  sold  43  %  of  his  stock  of  millinery  for 
$6875.35.  What  was  the  value  of  the  remainder  of  his  stock  of 
millinery? 

2.  A  and  B  are  partners,  A  having  invested  $19161  and  B 
$23419.  What  %  of  the  capital  of  the  firm  did  each  partner 
invest? 


146  PERCENTAGE 

3.  A  cistern  now  contains  741  gallons  of  water,  24  %  having 
leaked  out.     How  many  gallons  did  it  originally  contain? 

4.  A  clerk  spent  $676.50  of  his  annual  salary,  and  saved 
$523.50.     What  %  of  his  salary  did  he  spend? 

5.  If  A's  age  is  25  %  greater  than  B's,  what  %  of  A's  age  is 
B's? 

6.  A  is  worth  $45000  and  B  $30000.     What  %  more  than  B  is 
A  worth?    What  %  less  than  A  is  B  worth? 

7.  A  man  inherited  $74500,  invested  45  %  of  it  in  real  estate, 
25  %  of  it  in  United  States  bonds,  and  loaned  the  remainder  to  a 
friend.     What  was  the  sum  loaned? 

8.  A  bakery  used  425  pounds  of  flour  in  making  748  loaves  of 
bread,  each  loaf  weighing  12  ounces.     What  %  more  than  the 
flour  did  the  bread  weigh? 

9.  A  owns  12  %  of  the  stock  of  a  corporation,  B  15  %,  C  20  %, 
D  30  %,  and  E  the  remainder.     If  B's  investment  is  $3600,  what 
is  E's? 

10.  A  man  sold  40%  of  a  tract  of  land  at  one  time,  20%  of 
the  remainder  at  a  second  sale,  25  %  of  what  still  remained  at  a 
third  sale,  and  then  had  234  acres  finally  remaining.     How  many 
acres  did  the  original  tract  contain? 

11.  In  1910,  a  merchant's  sales  amounted  to  $185920,  and  in 
1911,  his  sales  were  $158032.     What  was  the  per  cent,  of  decrease 
in  1911? 

12.  A  certain  property  has  depreciated  $3375  in  value,  and 
is  now  worth  $9125.     What  was  the  %  of  depreciation? 

13.  A  man  bought  one  house  for  $4500  and  another  house 
for  42  %  more.    What  was  the  cost  of  both? 

14.  On  Sept.  23,  1911,  merchandise  was  bought  for  $1340  on 
3  months'  credit,  and  subject  to  a  discount  of  3  %  if  paid  within 
10  days.     What  is  the  proper  payment  on  Oct.  1,  1911? 

15.  A  firm's  business  was  placed  in  the  hands  of  a  receiver 
who  discovered  that  its  total  resources  after  deducting  the  costs 
of  receivership  were  $28500,  and  that  its  total  liabilities  were 
$37500.     What  %  of  the  firm's  indebtedness  can  the  receiver  pay 
to  the  creditors? 

16.  Two  railroads  carried  8260  pounds  freight  at  the  through 


PERCENTAGE  147 

rate  of  28  cents  per  cwt.  If  one  company  was  entitled  to  37  %  of 
the  through  charge,  how  much  should  the  other  company 
receive? 

17.  Two  railroads,  one  200  miles  long  and  the  other  300  miles 
long,  carried  750  barrels  flour  at  the  through  rate  of  18  cents  per 
barrel,  which  was  to  be  divided  pro  rata  between  them  (that  is,  in 
proportion  to  each  road's  per  cent,  of  the  total  mileage).     What 
%  of  the  total  freight  charge  should  each  road  have  received? 
What  percentage? 

18.  Three  railroads  respectively  200  miles,  250  miles,  and  350 
miles  in  length,  transported  28375  pounds  freight  at  the  through 
rate  of  28  cents  per  cwt.,  which  was  pro-rated  among  them  in  ac- 
cordance with  a  general  traffic  agreement.    What  %  of  the  total 
distance  did  each  road  carry  the  through  freight?     What  per- 
centage of  the  freight  collection  should  each  road  receive? 

19.  The  total  cost  of  an  arithmetic  and  a  geography  is  $2.10, 
and  the  arithmetic  cost  32  %  less  than  the  geography.    What  is 
the  cost  of  each? 

EXPLANATION.  The  required  cost  of  the  geography  is  indicated  as  the 
verbal  base  or  multiplicand  by  being  placed  immediately  after  "  %  less  than." 
$2.10  is  a  percentage  because  it  is  compared  with  (by  being  expressed  in  terms 
of)  the  indicated  base.  As  $2.10  is  the  joint  cost  of  the  geography  (the  indi- 
cated base)  plus  the  cost  of  the  arithmetic  (a  percentage),  its  specific  joint  rate 
must  be  the  specific  rate  of  the  geography  or  base  (100  %  of  the  cost  of  the 
geography)  plus  the  specific  rate  of  the  cost  of  the  arithmetic  or  percentage 
(100  %  -  32  %,  or  68  %  of  the  cost  of  the  geography),  that  is,  168  %  of  the  cost 
of  the  geography.  Hence,  divide  the  joint  percentage  or  product  ($2.10) 
by  its  joint  rate  %  or  multiplier  (168  %),  obtaining  $1.25  as  the  common  base 
or  multiplicand  (the  cost  of  the  geography). 

Then,  the  cost  of  both  ($2.10)  minus  the  cost  of  the  geography  ($1.25) 
should  equal  the  cost  of  the  arithmetic  ($.85).  Or,  the  obtained  common  base 
or  multiplicand  ($1.25)  multiplied  by  the  specific  rate  or  multiplier  for  the 
arithmetic  (68  %)  will  produce  the  specific  percentage  or  product  of  the 
arithmetic  ($.85). 

20.  A  man  bought  a  corner-house  and  the  house  adjoining  it 
for  $7752,  paying  28  %  more  for  the  corner-house  than  for  the  one 
adjoining  it.     How  much  did  he  pay  for  each? 

21.  A,  B,  C,  and  D  are  partners.     A's  investment  is  20% 
greater  than  B's,  C's  investment  is  10%  less  than  A's,  and  D's 


148  PERCENTAGE 

investment  is  75%  of  C's.     If  the  capital  of  the  firm  is  $53579, 
what  is  each  partner's  investment? 

NOTE.  If  two  or  more  bases  are  indicated  in  the  same  problem,  accept 
the  initial  or  root  base  as  the  base  of  the  solution,  and  regard  the  remaining 
derivative  bases  as  so  many  percentages  of  the  root  base.  There  can  be  no 
consolidation  of  two  or  more  specific  rates  or  multipliers  except  when  they  are 
expressed  upon  one  common  base  or  multiplicand  [Prin.  1,  18]. 

22.  A  has  25%  more  money  than  B,  B  28%  more  than  C, 
C  40  %  less  than  D,  and  D  25  %  as  much  as  E.     If  all  five  together 
have  $641.20,  how  much  has  each? 

23.  The  sales  of  a  firm  were  increased  20%  per  annum  for 
three  successive  years.     If  the  total  increase  during  the  three 
years  was  $36400,  what  were  the  sales  during  the  second  year? 

24.  The  gross  earnings  of  a  railroad  company  during  the  year 
1910  were  $6781345.30,  of  which  $2346785.90  was  expended  for 
wages,  $385465.20  for  repairs,  $275864.20  for  fuel,  and  $1753285.90 
for  fixed  charges.     What  %  of  the  gross  earnings  was  each  of  the 
above  charges  (to  five  decimal  places)? 

25.  A  commission  merchant  sold  10%  of  a  consignment  of 
wheat  at  one  sale,  30  %  of  the  remainder  at  a  second  sale,  40  %  of 
what  still  remained  at  a  third  sale,  and  then  had  1701  bushels 
remaining.     How  many  bushels  of  wheat  were  in  the  consignment? 

26.  A  owned  20%  of  the  capital  stock  of  a  steamboat  com- 
pany, and  sold  40  %  of  what  he  owned  for  $4800.     At  the  same 
appraisement,  what  was  the  value  of  the  entire  capital  stock? 

PROFIT  AND  LOSS 

209.  INTRODUCTION.     Profit   and   Loss    problems  are  such 
percentage  problems  as  have  special  reference  to  gains  or  losses 
in  selling.     While  differing  in  no  respect  from  other  percentage 
problems,  they  are  grouped  together  for  special  treatment  because 
of  their   relative   commercial   importance,    and   the   consequent 
necessity  for   acquiring  thorough   familiarity  with   the   general 
principles  of  percentage  as  applied  to  them,  and  for  cultivating 
special  skill  in  computing  them. 

210.  The  cost  is  the  sum  paid  for  the  thing  purchased. 

NOTE  1.  The  prime  cost  of  merchandise  is  the  first  cost,  that  is,  the  cost 
before  any  expense  connected  with  the  purchase,  such  as  freight  from  place 


PROFIT   AND   LOSS  149 

of  purchase,  drayage,  custom-house  charges,  commission  on  purchase,  insur- 
ance, storage  after  purchase,  etc.,  has  been  added.  The  gross  cost  is  the  last 
cost,  that  is,  the  first  cost  increased  by  all  expenses  incidental  to  the  purchase 
from  the  moment  of  purchase  to  the  moment  of  sale. 

NOTE  2.  When  the  term  cost  is  used  without  a  qualifying  word,  the  actual 
cost,  or  gross  cost,  is  always  understood. 

211.  The  selling  price  is  the  sum  received  for  the  thing  sold. 

NOTE  1.  The  gross  selling  price  of  merchandise  is  the  total  sum  received 
from  the  sale,  before  any  expense  connected  with  the  sale  has  been  deducted. 
The  net  selling  price  is  the  gross  selling  price  diminished  by  all  incidental 
expenses  incurred  after  the  moment  of  sale,  such  as  commission  on  sales,  dis- 
counts allowed,  free  delivery  charges,  etc. 

NOTE  2.  When  the  term  selling  price  is  used  without  a  modifying  adjec- 
tive, the  actual  or  net  selling  price  is  understood. 

212.  The  profit  or  gain  is  the  excess  of  the  net  selling  price 
above  the  gross  cost. 

213.  The  loss  is  the  deficiency  of  the  net  selling  price  below 
the  gross  cost. 

214.  Relation  of  terms.     1.   A  gross  cost  is  a  base  or  multi- 
plicand.   2.   A  %  of  profit,  of  loss,  or  of  selling  price,  is  a  rate  or 
multiplier.     3.   A  profit,  a  loss,  or  a  selling  price,  is  a  percentage 
or  product. 

NOTE  1.  Profit  means  increase  (more  than  cost)  and  loss  means  decrease 
(less  than  cost).  Calculations  in  profit  or  loss  are  therefore  applications  of 
207.  As  the  gross  cost  is  uniformly  increased  by  the  profit,  or  diminished  by 
the  loss  to  find  the  net  selling  price,  the  gross  cost  is  a  normal  base  or  standard 
of  comparison,  and  does  not  require  to  be  indicated  as  is  necessary  for  arbi- 
trary bases.  "Of  gross  cost"  is  therefore  understood  after  every  rate  of  profit 
or  of  loss,  when  no  other  base  is  arbitrarily  indicated. 

NOTE  2.  Since  the  selling  price  equals  the  cost  plus  the  profit,  or  the  cost 
minus  the  loss,  the  specific  rate  of  selling  price  must  equal  the  specific  rate  of 
cost  (100  %)  plus  the  specific  rate  of  profit,  or  minus  the  specific  rate  of  loss. 
[Note  1,  201.] 

EXAMPLES  FOR  PRACTISE 

RULE.  If  the  cost  (base)  and  a  rate  are  given,  multiply  the  cost  by  the 
specific  rate  of  the  required  profit,  or  loss,  or  selling  price. 

Goods  cost 

1.  $57.80  and  are  sold  at  25  %  profit.     Find  profit. 

2.  $183.75  and  are  sold  at  18  %  loss.     Find  loss. 

3.  $275.80  and  are  sold  at  28%  profit.     Find  profit. 


150  PERCENTAGE 

Goods  cost 

4.  $628.50  and  are  sold  at  20  %  loss.     Find  loss. 

5.  $342.16  and  are  sold  at  16f  %  profit.   Find   profit. 

6.  $486.25  and  are  sold  at  33J  %  loss.     Find  loss. 

7.  $254.85  and  are  sold  at  35  %  profit.     Find  selling  price. 

8.  $685.90  and  are  sold  at  25  %  profit.     Find  selling  price. 

9.  $816.35  and  are  sold  at  10  %  profit.     Find  selling  price. 

10.  $723.40  and  are  sold  at  5  %  loss.     Find  selling  price. 

11.  $374.85  and  are  sold  at  20  %  loss.  Find  selling  price. 

12.  $436.95  and  are  sold  at  12?  %  loss.  Find  selling  price. 

13.  $783.24  and  are  sold  at  45  %  gain.  Find  gain. 

14.  $294.35  and  are  sold  at  18  %  loss.  Find  loss. 

15.  $385.16  and  are  sold  at  24  %  gain.  Find  gain. 

16.  $416.28  and  are  sold  at  16  %  loss.  Find  loss. 

RULE.  If  the  cost  (base)  and  a  profit  or  loss  or  selling  price  (percentages) 
are  given,  divide  the  specific  percentage  or  product  of  which  it  is  required 
to  find  the  rate,  by  its  cost  or  multiplicand  factor,  and  the  quotient  carried  to 
hundredths,  will  be  its  specific  rate  %  or  multiplier  factor,  applying  a,  b,  or 
c,  205,  when  the  division  is  inexact. 
Goods  cost 

17.  $38.75  and  are  sold  at  $9.69  gain.     Find   %  of  gain. 

18.  $68.50  and  are  sold  at  $10.96  gain.     Find  %  of  gain. 

19.  $872.50  and  are  sold  at  $401.35  gain.     Find  %  of  gain. 

20.  $497.40  and  are  sold  at  $89.53  loss.     Find  %  of  loss. 

21.  $26.25  and  are  sold  at  $2.10  loss.     Find  %  of  loss. 

22.  $63.85  and  are  sold  at  $22.99  loss.     Find  %  of  loss. 

23.  $342.97  and  are  sold  for  $425.28.     Find    %  of  gain. 

24.  $682.25  and  are  sold  for  $777.77.     Find    %  of  gain. 

25.  $359.34  and  are  sold  for  $294.66.     Find  %  of  loss. 

26.  $458.37  and  are  sold  for  $401.07.     Find   %  of  loss. 

27.  $178.15  and  are  sold  for  $213.78.     Find  %  of  gain. 

28.  $728.35  and  are  sold  for  $591.78.     Find  %  of  loss. 

29.  $563.27  and  are  sold  at  $93.88  gain.     Find  %  of  gain. 

30.  $94.65  and  are  sold  at  $17.98  loss.     Find  %  of  loss. 

RULE.     To  find  the  cost,  divide  any  given  percentage  or  product  of  profit, 
of  loss,  or  of  selling  price,  by  its  specific  rate  or  multiplier  factor. 

31.  The  gain  is  $48.65  and  the  rate  of  gain  25%. 

32.  The  gain  is  $150.00  and  the  rate  of  gain  32  %. 


PROFIT   AND   LOSS  151 

Find  the  cost,  if 

33.  The  loss  is  $346.05  and  the  rate  of  loss  45  %. 

34.  The  loss  is  $168.65  and  the  rate  of  loss  20  %. 

35.  The  selling  price  is  $487.97  and  the  rate  of  gain  16f  %. 

36.  The  selling  price  is  $607.88  and  the  rate  of  loss  Yl\  %. 

37.  The  selling  price  is  $423.72  and  the  rate  of  gain  15%. 

38.  The  selling  price  is  $52.88  and  the  rate  of  loss  33£%. 

39.  The  selling  price  is  $348.70  and  the  rate  of  gain  37J  %. 

40.  The  selling  price  is  $58.65  and  the  rate  of  loss  6j  %. 

41.  The  selling  price  is  $105.19  and  the  rate  of  gain  34%. 

42.  The  selling  price  is  $406.50  and  the  rate  of  loss  35%. 

43.  The  selling  price  is  $519.86  and  the  rate  of  gain  10%. 
Find  gain. 

44.  The  selling  price  is  $130.58  and  the  rate  of  loss  15%. 
Find  loss. 

45.  Merchandise  was  sold  for  $48.40.     If  the  gain  was  $9.68, 
what  was  the  rate  %  of  gain? 

46.  Groceries  were  sold  for  $62.68.     If  the  loss  was  $15.67, 
what  was  the  rate  %  of  loss? 

47.  A  lot  of  hardware  was  sold  at  25%  gain.     If  the  seller 
received  $22.80,  how  much  of  it  was  profit? 

48.  A  house  was  sold  for  $2125  by  which  a  loss  of  15%  was 
sustained.     How  much  less  than  cost  was  received  by  the  seller? 

49.  A  man  bought  two  horses  at  $125  each.     He  afterwards 
sold  one  of  them  at  25  %  gain  and  the  other  at  15  %  gain.     How 
much  did  he  receive  for  the  two  horses? 

SUGGESTION.  A  common  cost  or  multiplicand  ($125)  multiplied  by  the 
specific  joint  rate  of  selling  price,  or  multiplier  for  the  joint  product 
(125  %  +  115  %)  =  joint  selling  price  or  product. 

60.  A  speculator  in  real  estate  bought  three  houses  at  $2400 
each,  and  afterwards  sold  one  of  them  at  a  profit  of  35%,  the 
second  at  a  profit  of  30  %,  and  the  third  at  a  profit  of  15  %.  How 
much  did  he  receive  for  the  three  houses? 

51.  Two  houses  were  sold  for  $2400  each,  by  which  the  seller 
gained  20  %  on  one  house  and  lost  20  %  on  the  other.  Did  the 
seller  receive  what  he  paid  for  the  two  houses?  If  not,  did  he 
gain  or  lose,  and  how  much? 


152  PERCENTAGE 

62.  Three  farms  were  bought  for  $3500  each,  and  they  were 
afterwards  sold  respectively  at  25%  gain,  30%  gain,  and  18% 
loss.  How  much  was  received  for  the  three  farms? 

53.  A  man  bought  a  tract  of  land  for  $4500,  paid  $625  addi- 
tional for  improvements,  and  then  sold  it  for  $6150.     What  was 
his  %  of  gain? 

54.  A  farmer  bought  48  acres  of  land  for  $3600,  and  afterwards 
sold  32  acres  from  his  purchase  at  a  profit  of  25%.     For  what 
sum  per  acre  must  he  sell  the  remainder  of  the  purchase  to  net 
an  average  profit  of  20  %  on  the  entire  purchase? 

55.  Allowing  for  a  loss  of  12  %  in  roasting,  what  should  be 
the  selling  price  of  roasted  coffee  which  was  bought  unroasted  at 
20  cents  per  pound,  to  net  a  gain  of  10%  [207]? 

56.  An  instalment  dealer  bought  furniture  at  $18  per  set. 
What  must  he  ask  per  set  to  net  a  profit  of  25  %,  after  allowing 
20  %  for  bad  collections? 

57.  A  wholesale  fruit  dealer  bought  a  vessel  load  of  water- 
melons numbering  12000  at  $16  per  hundred,  and  classifying  them 
according  to  size  and  quality,  sold  5000  of  them  at  $25  per  hun- 
dred, 3000  at  $20  per  hundred,  2500  at  $18  per  hundred,  1000  at 
$15  per  hundred,  and  the  remainder  for  $46.     What  was  his  % 
of  profit  on  the  entire  purchase? 

58.  A  merchant  sold  30  %  of  a  purchase  at  25  %  profit,  60  % 
of  the  remainder  at  20  %  profit,  and  what  still  remained  at  10  % 
profit.     What  was  his  %  of  gain  on  the  entire  purchase? 

59.  To  increase  their  capital,  a  firm  agreed  to  permit  one- 
half  of  the  profits  to  remain  hi  the  business  for  three  successive 
years.     The  net  gains  during  the  first  year  were  20  %,  during  the 
second  year  25%,  and  during  the  third  year  10%.     If  the  cap- 
ital of  the  firm  at  the  close  of  the  third  year  was  $31185,  what 
was  its  capital  at  the  beginning  of  the  first  year? 

60.  A  firm  sells  goods  in  its  wholesale  department  at  12% 
profit,  and  in  its  retail  department  at  25  %  advance  on  its  whole- 
sale selling  price.     What  %  profit  upon  original  cost  is  made  in 
the  retail  department? 

61.  If  20%  of  an  invoice  of  fruit  is  spoiled  in  shipment,  at 
what  %  gain  must  the  remainder  be  sold  to  net  10  %  profit  on  the 
original  cost? 


PROFIT  AND   LOSS  153 

62.  A  dealer  sold  merchandise  for  $48.99  and  gained  15%. 
What  should  have  been  the  selling  price  to  gain  25  %? 

63.  A  retail  grocer  bought  5  barrels  of  flour  at  $6.20  per 
barrel,  paid  25  cents  drayage  to  have  the  purchase  brought  to  his 
store,  and  afterwards  sold  one  barrel  to  a  customer  for  $7.75  and 
paid  25  cents  drayage  to  have  it  delivered  at  his  customer's  resi- 
dence.    What  was  the  grocer's  rate  %  of  profit? 

64.  If  the  sales  during  the  year  were  $60191,  the  inventory 
of  stock  on  hand  at  end  of  year  $9625,  and  the  gain  $7851,  what 
were  the  total  purchases,  and  the  %  of  gain? 

65.  If  the  purchases  during  the  year  are  $35798.56,  the  inven- 
tory of  stock  on  hand  at  end  of  year  $6500,  and  the  rate  of  gain 
12J  %,  what  are  the  total  sales? 

66.  The  books  of  a  manufacturer  for  the  year  1911  show  the 
following  costs:    productive  material  $200000,  productive  labor 
$300000,  overhead  expenses  $100000.      What  %  of  the  manufac- 
turing cost  is  each  element  of  expense?    [See  Suggestions  following.] 

SUGGESTIONS,  (a)  Manufacturing  or  factory  cost  consists  of  the  follow- 
ing elements:  productive  labor,  productive  material,  and  "overhead"  factory 
expense. 

(6)  Productive  or  direct  labor  is  labor  which  is  employed  directly  upon 
the  product  manufactured  to  sell.  The  labor  of  the  factory  superintendent 
foremen,  bookkeepers,  clerks,  porters,  etc.,  is  called  non-productive  or  in- 
direct labor. 

(c)  Productive  material  is  material  which  is  used  directly  in  the  articles 
manufactured  to  sell.     Material  used  for  general  purposes  around  the  factory, 
as  repairs,  etc.,  is  called  non-productive  material. 

(d)  Overhead  factory  expense  includes  non-productive  material;   non- 
productive labor;  and  any  general  factory  expense  as  heat,  light,   power, 
etc.,  except  such  as  is  incident  to  selling  the  manufactured  product. 

67.  In  the  manufacture  of  a  certain  make  of  bicycles,  the  pro- 
ductive labor  was  $82500,  productive  material  $71000,  and  over- 
head expense  $23000.     What  %  of  the  manufacturing  cost  is  the 
productive   labor?     Is   the   productive   material?     Is  the   over- 
head expense? 

68.  Factory  order  No.  16713  for  100  sewing  machines  shows 
the  following  costs:   direct  labor  $500,  productive  material  $600. 
Estimating  the  overhead  expense  at  40%  of  -the  direct  labor, 
what  was  the  manufacturing  cost  of  each  sewing  machine? 


154  PERCENTAGE 

NOTE  3.  In  manufacturing  cost  calculations,  though  not  the  uniform 
practise,  it  is  customary  to  distribute  the  indirect,  unproductive  or  overhead 
expense  among  the  units  of  product  at  an  estimated  %  upon  the  direct  labor 
employed  in  the  production  of  such  units. 

69.  In  a  certain  factory,  shop  order  No.  2519  shows  consump- 
tion of  productive  materials  $83.67,  direct  labor,  $41.83.     Esti- 
mating the  overhead  expense  at  40%  of  the  productive  labor, 
what  is  the  factory  cost? 

70.  Shop  order  No.   1627  of  Monumental  Machine  Works 
includes  the  following  costs:  direct  labor  $10248,  productive  mate- 
rial $3722,  estimated  overhead  factory  expense  25%  of  direct 
labor,  estimated  distributing  expense,  shipping,  etc.,  3  %  on  total 
sales.     What  is  the  selling  price  to  net  25  %  profit  on  factory  cost? 

71.  Estimating  50%  of  factory  cost  for  direct  labor,  what 
should  be  the  selling  price  of  75  units  of  product  at  16  f  %  profit, 
if  the  direct  labor  per  unit  is  $48? 

72.  The  annex  for  storing  the  raw  materials  of  a  factory 
with    its    contents    and    records    pertaining   to    the    same,    are 
destroyed  by  fire.      The  general  office  records  show   stores  on 
hand  at  last  inventory  $20000,  purchases  since  last  inventory 
$60000,  manufactured    and    sold    since    last  inventory   $96000, 
labor  and  material  on  unfinished  product  in  the  factory  at  the 
time  of  the  fire,  $14400.      The   manufacturing  costs  are  esti- 
mated at  50  %  for  direct  labor,  30  %  for  productive  material, 
and  the  remainder  for  overhead  expense.     If  the  manufacturer's 
selling    price  is  50%    advance   on  factory  cost,  what  was   the 
value  of  the  raw  material  in  the  annex  at  the  time  of  the  fire? 

.      TRADE  DISCOUNT 

215.  INTRODUCTION.    Problems  in  Trade  Discount  are  appli- 
cations of  the  general  principles  of  percentage  to  calculations  which 
have   special   reference   to   deductions   made  by   manufacturers 
and  jobbers  and  merchants  upon  sales  of  a  certain  magnitude 
to  stimulate  large  purchases;    or  as  a  compensation  to  middle- 
men for  handling  merchandise  to  sell  at  a  stipulated  list  price; 
or  as  an  absolute  allowance  to  purchasers  on   "time  prices"  if 
paid  for  at  any  time  within  a  given  period. 

216.  The  gross  price  is  the  price  before  any  deduction  has 
been  made. 


TRADE    DISCOUNT  155 

217.  The   trade   discount   is  the  deduction  which   has   been 
made  from  the  gross  price. 

218.  The   net   price  is  the  price  after   all  deductions   have 
been  made. 

219.  Identification  of  terms.     1.   A  gross  price  is  a  base  or 
multiplicand.     2.  A    %  of  trade  discount  or  of  net  price  is  a  rate 
or  multiplier.     3.  A  trade  discount  or  a  net  price  is  a  percentage 
or  product. 

NOTE  1.  Discount  (from  dis,  away;  count,  compute)  means  counting 
away.  In  taking  away  from  a  quantity,  it  is  decreased.  Hence,  problems  in 
trade  discount  are  applications  of  207,  relating  to  decreases;  and  the  quantity 
diminished  (the  gross  price)  is  the  normal  base  always  understood;  and  the 
trade  discount  (the  decrease),  and  the  net  price  (the  quantity  after  the  de- 
crease) are  percentages. 

NOTE  2.  If  two  or  more  rates  of  discount  are  allowed,  the  gross  price  is 
the  root  base  upon  which  the  first  rate  of  discount  is  computed;  and  the  pro- 
ceeds of  the  first  discount  should  be  taken  as  a  derivative  base  upon  which  the 
second  rate  of  discount  is  computed;  and  the  proceeds  of  the  second  discount 
should  be  regarded  as  a  second  derivative  base  upon  which  the  third  rate  of 
discount  is  computed;  etc.,  etc. 

GROSS  PRICE  GIVEN,   TO  FIND  THE  NET  PRICE 
ILLUSTRATIVE  EXAMPLE 

220.  The  gross  price  is  $278.56,  upon  which  a  discount   of 
28  %,  20  %,  and  3  %  is  allowed.     Find  the  net  price. 

SOLUTION  EXPLANATION.     Multiply 

$278.56,  gross  price  or  root  base  ^«J?^f?  uC°l  °r  ^  bT 

'  b          ^  ($278.56)  by  the  specific  rate 

•72,   (100%  —  28%)  Of  proceeds  of  the  first  dis- 

55712  count  (100  %  -28%  =  72  %) 

194992  to  ^nc*  tne  Proceeds  of  the 

first  discount    ($200.5632). 


$200.5632,  proceeds  of  1st  discount  Then  muitipiy  the  obtained 

40.1126,  20  %  or  J  of  $200.563  derivative  base  ($200.5632)  by 

$160.4506,  proceeds  of  2d  discount  the  second  rate  of  discount  <20 

4  R1V5    ^  <7  of  1*160  4^06  %==^  obtaming$40.1126,  and 

4^lt3°'  d  %  C  subtract,  leaving  $160.4506  as 

$155.6371,  proceeds  of  3d  discount  the  proceeds  of  the  second  dis- 

count. 

Then  multiply  the  last  obtained  derivative  base  ($160.4506)  by  the  third 
rate  of  discount  (3  %),  obtaining  $4.8135,  and  subtract,  leaving  $155.64  — 
as  the  proceeds  of  the  third  discount,  or  required  net  price. 


156  PERCENTAGE 

NOTE  1.  An  aliquot  rate  is  one  which  is  an  exact  divisor  of  100  %,  as  20  % 
in  the  111.  Ex.;  and  a  non-aliquot  rate  is  one  which  is  not  an  exact  divisor  of 
100  %,  as  28  %  in  the  111.  Ex.  Aliquot  rates  of  discount  are  more  conveniently 
handled  by  first  finding  the  discount  and  then  subtracting;  but  non-aliquot 
rates  of  discount,  by  at  once  multiplying  the  base  by  the  specific  rate  of  net 
price,  as  by  100  %  -  28  %,  or  72  %,  in  the  111.  Ex. 

NOTE  2.  A  rate  of  discount  consisting  of  a  single  figure,  as  3  %  in  the  111. 
Ex.,  is  best  computed  as  follows:  Commencing  with  the  cents'  order  of  the 
base,  multiply  it  and  each  of  its  successive  higher  orders  by  the  rate  of  discount, 
and  write  the  successive  orders  of  the  product  two  places  to  the  right  of  the 
orders  multiplied,  first  multiplying  the  rejected  orders  at  the  right  of  the 
cents'  order  of  the  base  to  obtain  the  correct  carrying  figure  to  the  retained 
product. 

NOTE  3.  All  results  should  be  carried  to  four  decimal  places,  to  insure 
accuracy  in  the  cents'  order  of  the  final  result. 

NOTE  4.  A  fraction  of  a  half  or  more  of  the  lowest  retained  order  is  not 
usually  counted  in  computing  a  discount  (which  is  against  the  seller),  and  is 
counted  as  another  unit  of  the  lowest  retained  order  in  computing  a  net  price 
(which  is  in  favor  of  the  seller).  Not  counting  in  the  former  instance  is 
equivalent  to  counting  in  the  latter. 


EXAMPLES  FOR  PRACTISE 

Find  the  net  price  Find  the  net  price 

Gross  Price        Rates  of  Disc't  Gross  Price  Rates  of  Disc't 

1.  $375.85 .  .  25  %,  10  %,  5  %  6.  $285.94 .  .  32  %,  24  %,  5  % 

2.  $582.90 . .  25  %,  20  %,  10  %  7.  $723.56 .  .  25  %,  16  %,  9  % 

3.  $873.62 . .  10  %,  10  %,  10  %  8.  $938.25 . .  16f  %,  12J  %,  2  % 

4.  $462.75 . .  20  %,  10  %,  7  %  9.  $427.36 . .  10  %,  7  %,  1  % 

5.  $629.38 . .  10  %,  5  %,  3  %  10.  $246.78 .  .  5  %,  3  %,  2  % 

11.  Goods  are  invoiced  at  $658.90  and  25  %,  5  %,  and  3  %  off. 
What  is  the  net  cost  to  the  purchaser? 

12.  Bought  2  doz.  pocket  knives,  No.  278,  at  $4.50  per  dozen; 
\  doz.  butcher  knives,  No.  715,  at  $7.50  per  dozen;   and  3  doz. 
locks,  No.  645,  at  $3.75  per  dozen.     What  was  the  net  cost, 
after  allowing  a  discount  of  20  %,  5  %,  and  2  %? 

13.  Bought  May  16,  1912,  a  bill  of  merchandise  amounting 
to  $738.65  gross  and  payable  in  90  days,  upon  which  a  merchan- 
dise discount  of  10  %,  10  %,  and  5  %  was  allowed,  with  a  time 
discount  of  3  %  if  paid  within  10  days.     What  should  be  the 
payment  if  made  in  full  on  May  25,  1912? 


TRADE    DISCOUNT  157 

REDUCTION  OF  RATES  OF  DISCOUNT 
ILLUSTRATIVE  EXAMPLES 

221.   1.  Reduce  the  discount  series  of  20%,  10%,  and  5% 
to  an  equivalent  single  discount. 

EXPLANATION.    Sub- 
SOLUTION 


100  %  =  gross  price  discount  (20%  of  the  gross 

20  %  =  1st  discount  price)   from  the  rate  of 

~80  %  =  proceeds  of  1st  discount    eross  price  (100  %)  to  find 
,  *       .  „„.  ~  ,    ,.  the  rate  of  1st  proceeds 

(TV  of  80)  _8  %  =  2d  discount  (8Q  %  of  the  ^  price) 

72  %  =  pro.  of  2d  discount  Next    subtract    the 

(?V  of  72)   3.6%  =  3d  discount  rate  of  the  2d  disc't  tfff 

6M  %  -  j  pro.  of  3d  discount        JJ^  J^  *£ 

{  or  rate  of  net  price         price)  from  the  pr0ceeds 

100  %  -  68.4  %  =  31.6  %,  single  discount      of  the  1st  disc't  (80  %  of 

the  gross  price)  to  find 
the  proceeds  of  the  2d  disc't  (72  %  of  the  gross  price). 

Next  subtract  the  rate  of  the  3d  disc't  GV  of  72  %  of  the  gross  price  or 
3.6  %  of  the  gross  price)  from  the  proceeds  of  the  2d  disc't  (72  %  of  the  gross 
price)  to  obtain  the  rate  of  proceeds  of  the  last  discount  (68.4  %  of  the  gross 
price). 

The  difference  between  100  %  of  the  gross  price  and  the  rate  of  final  pro- 
ceeds (68.4  %  of  the  gross  price),  or  31.6  %  of  the  gross  price  must  therefore 
be  the  total  of  the  three  discounts  expressed  in  a  required  single  discount  upon 
the  gross  price. 

2.  Reduce  the  discount  series  of  32  %,  24  %,  and  16  %  to  an 
equivalent  single  discount. 

SOLUTION  EXPLANATION.     If  32  %  of  the 
gross    price    (100  %)    is   taken   off, 

100  %  —  32  %  =  68  %  then  68  %  of   the  gross  price  will 

100  %  -  24  %  =  76  %  remain. 

100  %  —  16  %  =  84  %  If  24  %  of  what  remains  of  the 

68  X   76  X   84  =   434112  gross  price  is  tnen  taken  off>  76  % 

100  %  --  43.4112  %  -  56.5888  %  ° 


If  16  %  of  what  yet  remains  of  the  gross  price  is  finally  taken  off,  then 
84  %  of  76  %  of  68  %  of  the  gross  price,  or  43.4112  %  of  the  gross  price,  will 
finally  remain  as  the  rate  of  net  price. 

Then  subtract  the  rate  of  final  net  price  (43.4112  %)  from  the  rate  of 
the  original  gross  price  (100  %)  to  find  the  rate  of  total  discount  (56.5888  %). 


158  PERCENTAGE 

NOTE  1.  The  solution  of  111.  Ex.  1  is  preferable,  if  the  rates  of  discount 
are  aliquots  [Note  1,  220];  but  when  the  rates  of  discount  are  non-aliquots, 
the  solution  of  Ex.  2  will  be  the  more  convenient.  If  the  same  problem  con- 
tains both  classes  of  rates,  employ  the  method  of  Ex.  1  for  aliquot  rates,  and  of 
Ex.  2  for  the  non-aliquot  rates. 

Note  2.  The  value  of  a  series  of  discounts  will  not  be  changed  by  any 
variation  in  the  order  of  their  arrangement.  Thus,  a  discount  series  of  10  %, 
5  %  and  3  %  upon  any  gross  price  will  produce  the  same  net  price  as  5  %,  3  % 
and  10  %,  or  as  3  %,  10  %  and  5  %;  for  the  product  of  (100  —  10)  X  (100  -  5) 
X  (100  —  3)  will  be  the  same,  irrespective  of  the  order  in  which  they  may  be 
multiplied. 

EXAMPLES  FOR  PRACTISE 

Find  the  rate  of  net  price,  if  the  rates  of  discount  are 

1.  20  %,  10  %  4.  20  %,  20  %,  10  %     7.  35  %,  24  %,  8  % 

2.  25  %,  10  %,  5  %     5.  25  %,  20  %,  10  %     8.  20  %,  15  %,  10  %,  2  % 

3.  10%,5%,3%       6.  10%,10%,10%     9.  10%,5%,3%,1% 

Reduce  the  following  to  equivalent  single  discounts: 

10.  25  %,  10  %  12.  20  %,  5  %,  3  %       14.  20  %,  10  %,  10  % 

11.  20  %,  3  %  13.  10  %,  10  %,  5  %     15.  5  %,  3  %,  2  % 

TO  FIND  WHAT  PER  CENT.   OF  DISCOUNT  WILL  NET  A 
GIVEN  PRICE 

ILLUSTRATIVE  EXAMPLE 

222.  What  %  of  discount  must  be  allowed  to  net  $14.60,  if 
the  merchandise  is  marked  $18.25? 

SOLUTION  EXPLANATION.    The  solution  of  this  ex- 

Si  8.  25  =  gross  price  ample  is  an  application  of  202.     Hence, 

14.50  =  net  price  Divide  the  specific  percentage  of  which 

Q  «c  _.  rlio™iin+  it  is  required  to  find  the   %,  that  is,  the  per- 

o.OO   —   ui&uuunu 


<*Q  ACAH    •    <B:IQ  o*  -  one/     centage  or  product  of    discount   ($3.65)  by 
=  ZU  %     its  base  or  multiplicand  factor    (the  gross 
price,  $18.25)  to  find  the  specific  rate  of  discount  or  multiplier  factor  (20  %). 

NOTE.  If  the  division  is  not  exact,  as  is  highly  probable,  apply  a,  6, 
c,  205,  to  insignificant,  significant  and  intermediate  remainders;  and  apply 
Note,  206,  to  suggestive  rates. 

RULE.  Divide  the  specific  percentage  of  discount  (product) 
by  the  gross  price  (its  base  or  multiplicand  factor)  and  the  quotient 
carried  to  hundredths  will  be  the  rate  %  of  discount  (its  multiplier 
factor). 


TRADE    DISCOUNT  159 

EXAMPLES  FOR  PRACTISE 

What  %  of  discount  must  be  allowed  to  net 

1.  $34.01,  if  goods  are  marked  $35.80? 

2.  $60.32,  if  goods  are  marked  $75.40? 

3.  Cost,  if  goods  cost  $43.83  and  are  marked  $48.70? 

4.  Cost,  if  goods  cost  $58.80  and  are  marked  $67.20? 

5.  25  %  profit,  if  goods  cost  $75  and  are  marked  $133.93? 

6.  20  %  profit,  if  goods  cost  $85.35,  and  are  marked  $122.90? 

7.  5%  loss,  if  goods  cost  $15.80,  and  are  marked  $17.66? 

8.  Goods  are  marked  $38.25.     What  second  rate  of  discount 
must  be  used  with  20  %,  to  net  $29.07? 

THE  NET  PRICE  BEING  GIVEN,  TO  FIND  THE  GROSS  PRICE 
ILLUSTRATIVE  EXAMPLE 

223.   At  what  price  must  goods  be  marked  to  net  $5.25  after 
allowing  a  discount  of  25  %,  5  %,  and  3  %? 

SOLUTION  EXPLANATION. 

1.00          =  rate  of  gross  price       Divide   the  specific 

.25  =  rate  of  1st  discount    percentage   or   con- 

75          =  rate  of  1st  proceeds    tinued    P™f0c*   of 
,  i       f   --N  ^^r  e  r»  i    T  net  price  ($5.25)  by 

(A  of  .75)  .0375      =  rate  of  2d  discount     its  Pspecific  rate   >t 

.7125       =  rate  of  2d  proceeds  net  price  or  contin- 

(.7125  X  .03)  .021375  =  rate  of  3d  discount  ued    multiplier  fac- 

r  .  tor  (.691125),  found 

.691125  =  rate  of  net  price  M   ^  ^   221 

321  60  obtaining    $7.60    as 

.691125)$5.2500($7.p^  the  required  gross  or 

48379  marked     price,     its 

4191  multiplicand  factor. 

The  short  method 
for  dividing    by    a 

665  decimal    divisor    of 

g22  many  places  is  fully 

i     ^e    f   ™  explained  in  130. 

43  =  more  than  half  of  .69  The    gmall    in_ 

dex  figures  are  placed    over    the   lowest  decimal  order  of   the  successive 

i  2 

divisors;   that  is,  the  first  divisor  is  .6911,  the  second  .691,  the  third  and 

3 
last  .69. 


160 


PERCENTAGE 


RULE.  Divide  the  specific  percentage  of  net  price  (product)  by 
the  specific  rate  of  net  price  (its  multiplier  factor),  and  the  quotient 
will  be  the  gross  or  marked  price  (its  multiplicand  factor). 


EXAMPLES  FOR  PRACTISE 

At  what  price  must  goods  be  marked  to  net 

1.  $36.50  after  allowing  a  discount  of  25  %? 

2.  $18.75  after  allowing  a  discount  of  15  %  and  10%? 

3.  $62.50  after  allowing  a  discount  of  25  %,  10  %,  and  7  %? 

4.  $213.63  after  allowing  a  discount  of  25  %,  20  %,  5  %,  3  %? 

5.  $85.42  after  allowing  a  discount  of  30  %,  20  %,  10  %,  4  %? 

6.  25  %  profit?     Cost,  $24.36.     Discounts  10  %,  10  %,  5  %. 

7.  16|%  profit?    Cost,  $45.60.     Discts.,  20  %,  10  %,  3  %,  1  %. 

8.  10  %  loss?     Cost,  $16.85.     Discounts,  10  %,  10  %,  4  %,  2  %. 

9.  12J  %  loss?     Cost,  $9.60.     Discounts,  25  %,  5  %,  5  %,  3  %. 
10.   What  payment  on  May  28,  1912,  will  be  in  full  of  the 

following 


INVOICE 


BALTIMORE,  May  20,  1912. 


Mr.  R.  M.  BROWNING, 

Bought  of  F.  A.  SADLER  &  Co. 

Terms:  60  days,  net;  10  days,  2  %. 


3  doz 

Butcher  Knives  

$4.50 

** 

** 

25 

%,  20  %,  10  % 

* 

** 

* 

** 

$4.25  $5.75  I 

7.50 

2    " 

Pocket  Knives  ea.   345,     460, 

624  

** 

** 

10 

%,  10  %,    2  % 

* 

** 

** 

** 

$6.50  $7.25 

i     it 

Hatchets  ea      783      825 

* 

** 

$5.00  $6.25  $7.15 

5    " 

Spades  ea    No  1  No  2  No  3 

** 

** 

$4.50  $6.00 

3    " 

Shovels  ea.  No.  1  No.  2  

** 

** 

*** 

** 

50  %,  10  % 

** 

** 

** 

** 

** 

** 

Packing  and  Drayage 

50 

INTEREST 

224.  INTRODUCTION,  (a)  In  the  "Introduction"  of  193,  it 
was  shown  that  Percentage  is  an  application  of  the  general  prin- 
ciples which  govern  the  relation  of  all  quantities.  These  same 
general  principles  will  now  be  shown  to  apply  to  quantities  which 
include  a  hitherto  unconsidered  element,  that  of  time. 

(6)  Interest  is  a  compensation  for  the  use  of  money  in  the 
same  way  as  rent  is  a  compensation  for  the  use  of  property,  or 
wages  for  the  use  of  personal  labor.  In  all  of  these  it  is  evident 
that  the  amount  of  the  compensation  will  vary  with  the  length 
of  time  they  are  used.  Hence,  the  necessity  of  establishing  a 
fixed  unit  of  time  by  which  to  graduate  the  amount  of  compensa- 
tion, as  wages  at  so  many  dollars  per  day,  rent  at  so  many  dollars 
per  month,  and  interest  at  so  much  %  of  the  principal  per  year. 

(c)  It  is  thus  seen  that  the  element  of  time  is  not  a  new  term 
of  percentage,  but  a  necessary  modifier  of  one  of  the  three  already 
existent  terms  (the  rate)  which  is  no  longer  absolute  as  before, 
but  conditioned  upon  one  year's  use,  as  its  specific  name  "rate 
per  annum"  implies.  Hence,  at  5%  per  annum,  the  correspond- 
ing rate  for  two  years  will  be  two  times  5%,  or  10%;  for  three 
years  will  be  three  times  5%,  or  15%;  for  73  days  will  be  ?VV,  or 
i  of  5%,  or  1%;  for  146  days  will  be  ^|f  ,  or  §  of  5%,  or  2%,  etc. 


225.  Interest  is  a  compensation  for  the  use  of  money. 

226.  The  principal  is  the  sum  of  money  loaned,  or  the  sum 
of  money  due  but  unpaid,  for  which  a  compensation  is  charged. 

NOTE.  The  sum  of  the  principal  and  its  accrued  interest  is  called  the 
amount. 

227.  The  rate  of  interest  is  the  %  of  the  principal  which  is 
charged  for  using  it  one  year. 

NOTE.  The  specific  rate  of  interest  for  less  than  one  year  is  such  a  part 
of  the  rate  per  annum,  as  the  given  time  in  days  is  part  of  the  number  of  days 
in  a  year.  Thus,  counting  12  months  of  30  days  each,  or  360  days  to  a  year, 
as  is  ordinarily  done  in  interest  calculations,  at  6  %  per  annum,  it  will  be  i  of 
6  %,  or  1  %  for  £  of  360  days,  or  60  days;  and  if  1  %  of  the  principal  is  the  rate 
of  interest  for  60  days,  the  rate  of  interest  for  twice  60  days,  or  120  days  must 
be  twice  1  %,  or  2  %;  for  £  of  60  days,  or  15  days,  must  be  |  of  1  %,  or  \  %; 
and,  generally,  the  specific  rate  of  interest  for  any  number  of  days  at  6  %  per 
annum  is  such  a  part  of  1  %  (the  rate  of  interest  for  60  days),  as  the  given  time 
in  days  is  part  of  60  days;  or,  stated  in  a  form  for  convenient  use,  is  equal  to 
the  quotient  of  the  given  days,  regarded  as  hundredths,  divided  by  60.  The 


162  PERCENTAGE 

specific  rate  for  less  than  1  year  at  other  rates  per  annum  than  6  %  is  similarly 
obtained;  that  is,  the  specific  rate  of  interest  for  any  given  time  and  at  any  given 
rate  %  per  annum  must  equal  1  %  of  the  given  time  expressed  in  days  divided  by 
such  a  part  of  360  days  as  I  %  is  part  of  the  given  rate  per  annum.  Hence,  if 
the  specific  rate  of  interest  is  sought  for  216  days  at  5  %  per  annum,  divide  1  % 
of  216  da.  (2.16  da.)  by  i  of  360  da.  (72  da.)  obtaining  2.16  -=-  72,  or  .03;  to 
find  the  specific  rate  of  interest  for  225  da.  at  8  %  per  annum,  divide  1  %  of 
225  da.  (2.25  da.)  by  1  of  360  da.  (45  da.),  obtaining  2.25  -5-  45,  or  .05;  etc. 

228.  Simple   interest  is  a  compensation  for  the  use  of   the 
original  principal,  or  for  any  unpaid  part  of  the  original  principal. 

229.  Ordinary  interest  is   interest   computed  in  the  ordinary 
manner  for  less  than  one  year  upon  the  assumption  that  the 
entire  year  consists  of  12  months  of  30  days  each,  or  of  360  days. 

230.  Identification  of  terms.     1.  The  principal  is  the  base  or 
multiplicand.    2.  The  specific   %  for  the   given  time  [Note,  227], 
is  the  rate  or  multiplier.    3.  The  interest  or  the  amount  for  the 
given  time  is  the  percentage  or  product. 


PERCENTAGE  METHOD  FOR  COMPUTING  INTEREST 

AT  6% 

ILLUSTRATIVE  EXAMPLES 
231.   1.  Find  the  interest  of  $685.74  at  6%  for  193  days. 

EXPLANATION.     $685.74  is  a  normal  base  understood 
[206]  or  the  multiplicand  [230,  1];    6  %  is  the  rate  or  mul- 
$685.74        tiplier,  for  one  year's  interest,  but  not  the  specific  rate  for 
jgg        193  da.     At  6  %  per  annum  of  360  da.,  the  corresponding 
—  :  -        rate  will  be  £  of  6  %,  or  1  %,  for  £  of  360  da.,  or  60  da.     The 
205722        specific  rate  of  interest  for  193  da.  must  therefore  be  as 
617166          many  times  1  %  of  the  principal  as  60  da.  are  contained 
68574  times  in  193  da.,  or  W  of  1  %  of  the  principal.     As  W  of 

"^  of  the  PrinciPal  e(lual  *V<&  of  the  principal,  or  £  as  many 
thousandths  of  the  principal  as  there  are  days,  that  is,  |  of 


$22.057  193  thousandths  of  the  principal,  therefore, 

Multiply  the  principal  or  multiplicand   ($685.74)  by 

6  times  the  specific  rate  or  multiplier  (.193),  obtaining  6  times  the  specific 
percentage  of  interest  or  product  ($132.34782);  and  divide  this  interest  by 
6  to  find  the  correct  interest  ($22.06-). 

2.   Find  the  interest  of  $417.28  for  2  yr.  228  da.  at  6  % 


INTEREST  163 

SOLUTION 

2  yr.  of  360  da.     -  720  da.  ExpLANATION.    There  are  twice  360  da. 

Fraction  of  a  yr.  =  228  da.     or  720  da.  in  the  2  complete  years,  and 
6). 948  228  da.  in  the  fraction  of  a  year,  or  720  da. 

Q       ./.  i  rQ  +  228  da.  (948  da.)   in  the  given  time. 

SPe'  Hence,  *  of  .948  of  the  principal,  or  .158 

of  the  principal  must  be  the  specific  rate 

Principal  =  $417.28  of  interest  for  the  given  time.     Therefore, 

Specific  rate   =         .158  Multiply  the  principal  or  multiplicand 

ooo  on  A  ($417.28)  by  the  specific  rate  of  interest 

**  or  multiplier  for  2  yr.  228  da.  (.158)  to  find 

the    specific    percentage   of    interest    or 

41728  product  ($65.93  +). 

$65.93024 

RULE.  Multiply  the  principal  by  one-thousandth  of  the  time 
expressed  in  days,  and  divide  the  product  by  6. 

NOTE  1.  If  the  days  are  exactly  divisible  by  6,  as  in  111.  Ex.  2,  it  will  be 
found  more  convenient  to  multiply  the  principal  by  one-sixth  of  the  days 
expressed  in  thousandths,  thus  minifying  the  multiplication  and  altogether 
avoiding  the  division  by  6. 

NOTE  2.  The  interest  at  any  higher  or  lower  rate  than  6  %  may  be  found 
by  proportionally  increasing  or  diminishing  the  interest  obtained  by  the  pre- 
ceding process.  Thus,  as  7  %  is  I  of  6  %,  so  the  interest  at  7  %  must  be  I  of 
the  interest  at  6  %;  as  5  %  is  |  of  6  %,  so  the  interest  at  5  %  must  be  |  of  the 
interest  at  6  %;  etc. 

NOTE  3.  If  the  time  is  expressed  in  months  and  days,  or  years,  months 
and  days,  reduce  it  to  days  by  allowing  360  days  to  each  year  and  30  days  to 
each  month.  Though  the  ordinary  interest  for  the  exact  time  between  two 
dates  is  required,  if  that  interval  contains  one  or  more  years,  each  year  should 
uniformly  be  regarded  as  containing  360  days. 


EXAMPLES  FOR  PRACTISE 

At  6  %  per  annum,  find  the  interest  of 

1.  $748.35  for  48  da.  6.  $613.28  for  324  da. 

2.  $813.72  for  174  da.  7.  $924.13  for  7  mo.  23  da. 

3.  $258.37  for  253  da.  8.  $416.25  for  5  mo.  16  da. 

4.  $521.63  for  217  da.  9.  $2358.16  for  1  yr.  8  mo.  13  da. 

5.  $1253.18  for  119  da.  10.  $758.30  for  3  yr.  10  mo.  24  da. 


164  PERCENTAGE 

PERCENTAGE  METHOD  FOR  COMPUTING  INTEREST  AT 
ANY  PER  CENT 

ILLUSTRATIVE  EXAMPLE 
232.   Find  the  interest  of  $675.83  at  5%  for  223  days. 

SOLUTION  EXPLANATION.     $675.83  is  the  principal  or  base  (mul- 

tiplicand) ;   and  5  %  is  the  rate  or  multiplier  for  one  year, 
$b75.8o        kut  no^  the  specific  rate  of  interest  or  multiplier  for  223  da. 
At  5  %  per  annum  of  360  da.,  the  corresponding  rate 
202749       must  be  *  °f  5  %»  or  1  %>  for  *  °f  3GO  da->  or  72  da-    There- 
13^1fifi          ^ore  *ke  sPec^c  rate  °f  Merest  or  multiplier  for  223  da. 
must  be  as  many  times  1  %  of  the  principal  as  72  da.  are 
135166  contained  times  in  223  da.,  or  W  of  1  %  of  the  principal. 

8)1507.1009       .As  -W  of  Tfor  of  the  principal  equal  fffo  of  the  principal, 
^r  7^  M  many  hundredths  of  the  principal  as  there  are  days, 


Q\-IOO 

that  is,  A  of  223  hundredths  of  the  principal,  therefore, 
$20.9319  Multiply  the  principal  or  multiplicand   ($675.83)  by 

72  times  the  specific  rate  or  multiplier  (2.23),  obtaining  72 
times  the  specific  percentage  of  interest  or  product  ($1507.1009);  and 
divide  by  72  (=  8  X  9)  to  find  the  correct  interest  ($20.93  +). 

RULE.  Multiply  the  principal  by  one-hundredth  of  the  time 
expressed  in  days,  and  divide  the  product  by  such  a  part  of  360  as 
1%  is  of  the  given  rate  per  annum. 

NOTE  1.  In  finding  the  interest  by  the  above  method,  it  will  materially 
lessen  the  mechanical  labor  to  cancel  any  factor  which  is  common  to  both 
multiplier  and  divisor.  Thus,  to  find  the  interest  at  8  %  for  63  days,  instead 
of  multiplying  the  principal  by  .63  and  dividing  by  45  (360  -J-  8),  it  will  be 
easier  to  multiply  by  $  of  .63  or  .07,  and  to  divide  by  i  of  45,  or  5. 

NOTE  2.  If  the  given  rate  per  annum  is  not  an  exact  divisor  of  360,  first 
find  the  interest  at  6  %,  and  apply  Note  2,  231. 

EXAMPLES  FOR  PRACTISE 

Find  the  interest  of  Find  the  interest  of 

1.  $568.73  at  8  %  for  97  da.  8.  $372.50  at  4  \  %  for  280  da. 

2.  $842.95  at  9  %  for  265  da.  9.  $563.75  at  3-|  %  for  297  da. 

3.  $673.48  at  5%  for  173  da.  10.  $728.34  at  7  %  for  264  da. 

4.  $498.25  at  4  %  for  82  da.  11.  $815.93  at  8  %  for  7  mo.  24  da. 
6.  $725.38  at  3  %  for  141  da.  12.  $425.30  at  5  %  for  5  mo.  12  da. 

6.  $284.83  at  10  %for  327  da.  13.  $785.90  at  4  %  for  1  yr.  279  da. 

7.  $916.40atl2%for285da.  14.  $893.25  at  3  %  for  3  yr.  175  da. 


INTEREST  165 

TO  COMPUTE  INTEREST  AT  6%  BY  "THE  ALIQUOT 
METHOD" 

233.  INTRODUCTION.  The  following  method  is  a  modification 
of  the  aliquot  method  in  common  use  which  will  enable  the  cal- 
culator to  verify  the  mechanical  accuracy  of  the  final  result  with- 
out reviewing  all  the  figures  used  in  obtaining  that  result.  By 
this  method,  the  aliquots  are  so  arranged  that  each  successive 
aliquot  is  an  aliquot  part  of  the  aliquot  immediately  preceding  it, 
and  that  the  last  aliquot  is  also  an  aliquot  part  of  the  initial  state- 
ment. This  arrangement  will  cause  the  verification  of  the  few 
figures  in  the  last  aliquot  to  include  the  verification  of  all  the  pre- 
ceding aliquots;  for  if  the  last  aliquot,  first  obtained  by  an  un- 
broken succession  of  derivations  from  the  initial  statement,  is 
found  to  be  correct  by  being  obtainable  by  direct  derivation  from 
the  same  initial  statement,  then  the  succession  of  aliquots  from 
which  it  was  originally  derived  must  also  be  correct. 

ILLUSTRATIVE  EXAMPLES 
1.   Find  the  interest  of  $283.75  at  6  %  for  210  days. 

SOLUTION  EXPLANATION.     At  the  rate  of  6  % 

I  •        r  P61*  annum  °f  360  da.,  the  correspond- 

Q  days     ing  rate  must  be  |  of  6  %,  or  1  %  of 

8 
1 


51  25  =     "      "    180      "          the  principal  for  £  of  360  da.,  or  60  da. 
41  g7  _     ((      n      30      tt  Therefore,  by  means  of  an  elon- 

-^"^ u 7T~1^  ~^~~      8ated  decimal  P°int>  cut  off  two  figures 

J  |  yd  1Z  -  J1U  from  the  right  of  the  dollars,  which  is 

equivalent  to  taking  1  %  of  the  prin- 
cipal [Prin.  5,  120],  thus  obtaining  the  initial  interest  for  60  da.  (Note, 
227.) 

Next  find  the  interest  for  the  greatest  exact  dividend  of  60  da.  (180  da.) 
which  is  found  in  the  given  number  of  days  (210);  and  as  180  da.  are  3  times 
60  da.,  the  interest  for  180  da.  must  be  3  times  the  interest  for  60  da.  ($2.8375 
X  3),  or  $8.5125. 

Next  find  the  interest  for  the  greatest  aliquot  part  of  180  da.  (the  last 
obtained  result)  which  can  be  found  in  the  remaining  30  da.  of  the  given  time 
(210  da.).  As  30  is  itself  an  aliquot  part  of  180  (30  =  |  of  180),  therefore  £ 
of  the  interest  for  180  da.  (J  of  $8.5125),  of  $1.4187,  must  be  the  interest  for 
30  da. 

The  interest  for  30  da.  added  to  the  interest  for  180  da.  must  equal  the 
interest  for  210  da.  ($9.93  +). 

VERIFICATION.  Observe  if  the  last  aliquot  ($1.4187),  supposed  to  be  the 
interest  for  30  da.,  is  one-half  of  the  initial  statement  ($2.8375)  which  is  known 


166  PERCENTAGE 

to  be  the  interest  for  60  da.  If  so,  it  should  be  accepted  as  correct,  as  the  last 
aliquot  ($1.4187)  has  now  been  obtained  from  two  independent  sources.  If 
the  interest  for  30  da.  is  correct,  then  the  interest  for  180  da.  must  also  be 
correct,  as  the  former  was  originally  derived  from  the  latter.  As  a  further 
precaution,  add  the  two  parts  of  the  time  (180  +  30)  to  see  if  they  equal  the 
entire  time  (210  da.).  Lastly  prove  the  addition  by  Note  1,  19. 

2.   Find  the  interest  of  $875.93  at  6  %  for  4  yr.  285  da. 

EXPLANATION.     If,  as  shown 
in  Ex.  1,  $8.7593  is  the  interest 

$8  |  75.93      =  int.  for    60  da.  of  $875.93  for  60  da.,  then  the 

interest  of  $875.93  for  4  times  60 


35 
5 


03  72    =  '   240 

83  95    =    "     "     40 


da.,  or  240  da.  (the  greatest  div- 
idend of  60  da.  found  in  285  da.) 


72  99     =  5    '  must  be    4    times    $8.7593, 

03  72  )  $35.0372. 


or 


72  ) 
Q    I  ~ 


35 

175  18  6  i  *  ^r*  ^  *ke  m*;erest  f°r  240  da.  is 

$35.0372,  the  interest  for  £  of 
$251  |  82  98  =  4  yr.  285  da.  240  da.,  or  40  da.  (the  greatest 

aliquot  part  of  the  immediately 

preceding  240  da.  which  is  found  in  the  remaining  45  da.  of  the  given  285 
da.)  must  be  i  of  $35.0372,  or  $5.8395. 

If  the  interest  for  40  da.  is  $5.8395,  the  interest  for  |  of  40  da.,  or  5  da. 
(the  greatest  aliquot  part  of  the  immediately  preceding  40  da.,  which  is  found 
in  the  remaining  5  da.  of  the  originally  given  285  da.)  must  be  |  of  $5.8395, 
or  $.7299. 

If  6  %  is  the  rate  of  interest  for  1  yr.,  4  times  6  %,  or  24  %  will  be  the  cor- 
responding rate  for  4  yr.  As  the  initial  statement  ($8.7593)  is  1  %  of  the 
principal,  24  times  $8.7593  ( =  4  times  the  initial  statement,  or  $35.0372;  and 
20  times  the  initial  statement,  or  $175.186)  must  be  the  interest  for  4  years. 

The  sum  of  the  interest  for  these  several  parts  of  the  full  time  is  therefore 
the  full  required  interest. 

VERIFICATION.  As  a  precaution  against  a  possible  mechanical  error 
observe  if  the  interest  for  5  da.  ($.7299)  is  12  of  the  initial  interest  for  60  da. 
($8.7593).  If  it  is,  not  only  is  it  correct,  but  also  all  the  preceding  aliquots; 
for  the  interest  for  5  days  was  originally  obtained  by  successive  derivations 
through  each  of  the  preceding  aliquots,  and  a  correct  result  cannot  be  derived 
from  an  incorrect. 

RULE.  1.  Cut  off  two  figures  from  the  right  of  the  dollars  in 
the  principal  by  an  elongated  decimal  point,  and  regard  the  result 
as  the  interest  for  60  days. 

2.  Take  as  many  times  the  interest  for  60  days  as  the  greatest 
dividend  of  60  days  in  the  given  days,  is  times  60  days. 


INTEREST  167 

3.  Take  such  an  aliquot  part  of  the  last  obtained  interest  as  will 
equal  the  greatest  aliquot  part  of  the  last  obtained  days  which  is 
found  in  the  remaining  days  of  the  given  time. 

4.  Thus  continue  until  the  required  interest  for  all  the  days  of 
the  given  time  has  been  found. 

NOTE  1.  If  care  is  taken  to  have  the  terminating  aliquot  days  also  an 
aliquot  of  60  da.,  it  will  only  be  necessary  to  test  the  last  aliquot  interest  by 
comparison  with  the  initial  interest  for  60  days,  to  verify  not  only  the  last 
result,  but  also  all  the  intermediate  aliquots  between  it  and  the  initial  state- 
ment. 

NOTE  2.  If  the  given  time  contains  years  and  days,  first  find  the  interest 
for  the  days,  and  verify.  Then  take  as  many  times  6  times  the  initial  state- 
ment (1  %  of  the  principal)  as  the  given  years  are  times  1  year,  and  verify. 
At  6  %  the  interest  for  years  is  most  conclusively  verified  by  comparing  one 
partial  product  with  the  other.  Thus,  the  multiplier  for  4  yr.,  is  4  times  6  %, 
or  24  %,  hence,  if  the  partial  product  by  4,  when  obtained,  should  be  twice 
the  partial  product  by  2,  it  will  prove  both  partial  products  to  be  correct. 

NOTE  3.  Should  any  given  interval  in  days  lack  an  aliquot  part  of  60  da. 
of  being  an  exact  dividend  of  60  da.,  it  will  usually  be  more  convenient  to 
increase  the  multiplier  by  1,  and  deduct  the  aliquot  excess  of  interest  from 
the  resulting  product.  Thus,  to  find  the  interest  for  117  da.,  first  find  the 
interest  for  120  da.  and  diminish  the  result  by  the  interest  for  3  da.  (*V  of  120, 
that  is,  I  of  the  interest  for  120  da.  written  one  place  to  the  right  of  the  orders 
divided). 

EXAMPLES  FOR  PRACTISE 
Verifying  results,  find  the  interest  at  6  %. 

Principal        Time  Principal             Time 

1.  $648.15    240  da.  10.  $425.93  204  da. 

2.  $375.34     186  "  11.  $537.52  15 

3.  $827.43     222   "  12.  $875.40  262   " 

4.  $516.20    345  "  13.  $647.52  25  " 

5.  $285.95     168  "  14.  $9246.30  146  " 

6.  $416.62      85  "  16.  $17893  78  " 

7.  $742.85     160  "  16.  $9.50  50  " 

8.  $924.50    275   "  17.  $68.75  2  yr.  153  da. 

9.  $618.80      42  "  18.  $87.25  7  yr.  311  da. 

NOTE  4.  By  compound  subtraction,  the  interval  between  two  dates  is  most 
conveniently  found  as  follows:  From  any  date  in  the  fifth  month  (say,  May 
19,  1912)  to  the  same  date  in  the  eleventh  month  (Nov.  19,  1912)  are  6  mo.  of 


168  PERCENTAGE 

30  da.  each  or  180  da.,  therefore  to  Nov.  27,  or  8  da.  later,  the  time  will  be  8  da. 
more  than  180,  or  188  da.,  and  to  Nov.  13,  or  6  da.  earlier  the  time  will  be  6  da. 
less  than  180,  or  174  da.  If  the  interval  is  greater  than  one  year,  first  find 
the  number  of  complete  years  in  the  given  interval,  and  then  as  above  described, 
find  the  time  in  the  remaining  fractional  part  of  a  year.  Thus,  from  Sept.  25, 
1908,  to  July  18,  1912,  are  3  yr.  293  da.;  that  is,  from  Sept.  25,  1908,  to  the 
latest  corresponding  date  in  the  interval  (Sept.  25,  1911)  are  3  years,  and  from 
Sept.  25,  1911  (9th  mo.  of  1911)  to  July  25,  1912  (7th  mo.  of  1912,  or  the 
equivalent  of  7  +  12,  or  the  19th  mo.  of  1911)  are  10  mo.  of  30  da.  each,  or  300 
da.;  and  to  July  18,  1912,  or  7  da.  earlier  than  July  25,  1912,  are  7  da.  less 
than  300  da.,  or  293  da. 

NOTE  5.  To  find  the  exact  number  of  days  in  any  fractional  part  of  a 
year,  first  find  the  time  by  the  30-day  method  [Note  4,]  and  add  one  day  for 
each  31-da.  mo.  in  the  given  interval  and  subtract  two  days  if  the  interval 
includes  the  last  day  of  February  of  a  common  year,  or  subtract  one  day  if  it 
includes  the  last  day  of  February  of  a  leap  year. 

Finding  the  time  by  Note  4,  what  is  the  interest  at  6  % 

19.  Of  $129.60  from  Sept.  18,  1910,  to  June  16,  1911? 

20.  Of  $673.45  from  Mch.  23,  1908,  to  Nov.  13,  1912? 

21.  Of  $827.64  from  Dec.  12,  1905,  to  May  6,  1911? 

22.  Of  $285.96  from  Oct.  31,  1903,  to  June  17,  1911?     [Note 
2,  163.] 

23.  Of  $875.49  from  May  13,  1905,  to  Oct.  31,  1911?     [Note 
2,  163.] 

24.  Of  $5168.35  from  Nov.  27,  1902,  to  July  12,  1912? 

25.  Of  $87.90  from  July  31,  1907,  to  Oct.  19,  1911?     [Note 
2,  163.] 

26.  Of  $8.53  from  Jan.  16,  1906,  to  Nov.  25,  1911? 

27.  Of  $62.70  from  June  30,  1908,  to  Feb.  29,  1912?     [Note 
2,  163.] 

28.  Of  $85378.15  from  Feb.  21,  1908,  to  May  17,  1912? 

29.  Of  $73.92  from  May  18,  1909,  to  Sept.  3,  1911? 

30.  Of  $2354.70  from  June  28,  1910,  to  Jan.  31,  1912? 

31.  Of  $825.37  from  Dec.  28,  1902,  to  Nov.  18,  1911? 

32.  Of  $497.85  from  Apr.  24,  1907,  to  July  18,  1911? 

33.  Of  $745.93  from  Aug.  18,  1905,  to  Sept.  24,  1912? 

Finding  the  exact  time  by  Note  5,  what  is  the  interest  at  6  % 

34.  Of  $978.42  from  May  23,  1911,  to  Dec.  16,  1911? 

35.  Of  $427.65  from  Oct.  15,  1911,  to  Mch.  31,  1912? 


INTEREST  169 

36.  Of  $695.80  from  Feb.  3,  1911,  to  Nov.  6,  1911? 

37.  Of  $584.75  from  June  30,  1911,  to  Aug.  5,  1911? 

38.  Of  $752.35  from  Dec.  15,  1911,  to  Dec.  22,  1911? 

39.  Of  $859.40  from  Jan.  20,  1908,  to  Nov.  26,  1908? 

40.  Of  $348.59  from  May  1,  1910,  to  Sept.  20,  1910? 

41.  Of  $2853.60  from  July  3,  1911,  to  Feb.  6,  1912? 

42.  Of  $695.73  from  Mch.  28,  1910,  to  Oct.  12,  1910?  ' 

43.  Of  $196.82  from  Nov.  17,  1910,  to  Apr.  5,  1911? 

44.  Of  $378.45  from  Apr.  12,  1911,  to  Aug.  24,  1911? 

45.  Of  $518.62  from  Dec.  19,  1910,  to  Feb.  23,  1911? 

46.  Of  $35.76  from  Oct.  8,  1911,  to  May  23,  1912? 

47.  Of  $9.20  from  Sept.  29,  1907,  to  Aug.  6,  1908? 

48.  Of  $18275.65  from  Aug.  13,  1909,  to  Jan.  25,  1910? 

49.  Of  $598.25  from  May  16,  1904,  to  Nov.  12,  1909? 

60.  Of  $643.50  from  July  28,  1908,  to  Mch.  3,  1912? 

61.  Of  $2468.75  from  June  24,  1905,  to  Apr.  6,  1911? 

62.  Of  $525.12  from  Feb.  25,  1908,  to  Aug.  31,  1911? 

63.  Of  $716.53  from  Dec.  19,  1903,  to  May  25,  1908? 


TO  COMPUTE  INTEREST  AT  ANY  PER  CENT.  BY  THE 
6%   METHOD 

ILLUSTRATIVE  EXAMPLE 
234.   Find  the  interest  of  $758.35  at  9%  for  215  days? 

SOLUTION  EXPLANATION.      First    find 

$7  |  58.35    =  int.  at  6  %  for    60  da.  the  interest  of   the  given  prin- 

u    i  QT\    u     cipal  for  the  given  time  at  6  % 


22 
3 


27 
13 


t 


79  17  =    "     "    "     "     30 


by  233,  obtaining  $27.1741. 
As  9  %  is  one-half  of  6  %  (3  %) 


63   19  =  5    '       added  to  6  %,  so  the  interest  at 


17  41  =     "      "  Qo/0    "    215    "     9  %  must  be  one-half  of  the  in- 

58  70=     "      "  3°7    "    215    "     terest  at  6  %  ($13.5870)  added  to 

the  interest  at  6  %   ($27.1741), 


$40  I  76  11  -          "  9%   "   215  '     or  $40.76+. 

RULE.  First  find  the  interest  at  6%,  then  regard  whatever  is 
over  or  under  6  %  in  the  given  rate,  as  an  aliquot  of  6  %,  and  increase 
or  diminish  the  interest  at  6%  by  the  corresponding  aliquot  part  of 
the  interest  at  6%. 


170  PERCENTAGE 

NOTE.      If  the  given  rate  is  more  than  one  aliquot  of  6  %  over  or  under 
6  %,  employ  Note  2,  231. 

EXAMPLES  FOR  PRACTISE 

Verifying  each  result,  find  the  interest  of 

1.  $578.25  at  7  %  for  165  da.     6.   $485.60  at  4  %  for  143  da. 

2.  $297.62  at  9%  for  273  da.     7.   $816.12  at  8%  for  85  da. 

3.  $815,40  at  5%  for  137  da.     8.   $3487. 25  at  4J  %  for  118  da. 

4.  $765.53  at  10%  for  315  da.     9.   $1852.70  at  3J  %  for  256  da. 

5.  $654.25  at  3  %  for  198  da.   10.    $96.85  at  5J  %  for  174  da. 

11.  $5.90  at  7J  %  for  1  yr.  216  da. 

12.  $816.35  at  12%  for  4  yr.  297  da. 


DIRECT  ALIQUOT  METHOD  OF  FINDING  THE  INTEREST 
AT  ANY  PER  CENT. 

ILLUSTRATIVE    EXAMPLE 
235.   Find  the  interest  of  $647.28  at  5  %  for  205  days. 

EXPLANATION.     At  the  rate 
SOLUTION  of    5%    of    the    principal    per 

$6  I  47.28   =  int.  at  5  %  for    72  da.  annum  of  36°  <**•>  the  c,°rre- 


/L/  1  •  j  •       1_  1  *     M 

spending  rate  must  be  -k  of  5 

12      9456=     "      "     "      "144"      of  the  principal,  or  1  %  of  the 

4      31   52  =     "      "     "      "      48    "      principal  for  &  of  360  da.,  or 

1      07  88  =     "      "     "      "      12    "      72  days.     Therefore,  by  means 

g  99  =     n      «     t(      u         i    (t      of  an  elongated  decimal  point, 

cut  off  two  orders  from  the  right 


$18  |  42  95  =  int.  at  5  %  for  205  da.  Of  the  dollars,  which  is  equiva- 
lent to  taking  1  %  of  the  principal 

[Prin.  5,  120]  and  thus  to  obtaining  the  interest  for  72  da.  at  5  %  per  annum 
[Note,  227]. 

Next  find  the  interest  for  the  greatest  dividend  of  72  da.  (144  da.)  which  is 
found  in  the  given  number  of  days  (205  da.) ;  and  as  144  da.  are  2  times  72  da., 
the  interest  for  144  da.  must  be  2  times  the  interest  for  72  da.  ($6.4728),  or 
$12.9456. 

Next  find  the  interest  for  the  greatest  aliquot  part  of  144  da.  (the  last 
obtained  result)  which  can  be  found  in  the  remaining  61  da.  of  the  given  time, 
or  for  48  da.  ( J  of  144  da.).  If  $12.9456  is  the  interest  for  144  da.,  |  of  $12.9456 
or  $4.3152  must  be  the  interest  for  £  of  144  da.,  or  48  da. 

Next  find  the  interest  for  the  greatest  aliquot  part  of  48  da.  (the  last 
obtained  result)  which  can  be  found  in  the  remaining  13  da.  of  the  given  time, 


INTEREST  171 

or  for  12  da.  ( £  of  48  da.) .  If  $4.3152  is  the  interest  for  48  da.,  i  of  $4.3152,  or 
$1.0788,  must  be  the  interest  for  |  of  48  da.,  or  12  da. 

Next  find  the  interest  for  the  remaining  1  da.  of  the  given  time,  which  is  i1? 
of  the  last  obtained  interest  for  12  da.,  or  $0.899. 

Lastly  add  these  several  partial  interest  results,  obtaining  $18.43-  as  the 
complete  interest. 

VERIFICATION.  Observe  if  the  last  aliquot  ($.0899)  is  7V  ( =  1  of  i)  of  the 
initial  statement  ($6.4728).  If  so,  the  last  aliquot  must  be  correct,  and  con- 
sequently all  the  preceding  aliquots  through  which  the  last  aliquot  was  suc- 
cos.-ively  derived,  must  also  be  correct.  As  a  further  precaution,  add  the 
aliquot  days  to  see  if  they  equal  205;  and  prove  the  final  addition  by 
Note  1,  19. 

RULE.  1.  //  360  be  divided  by  as  many  times  1  as  the  given 
rate  is  times  1  %,  the  quotient  will  be  the  initial  number  of  days  for 
which  1%  of  the  principal  is  the  initial  interest. 

2.  Take  as  many  times  the  initial  interest  as  the  greatest  exact 
dividend  of  the  initial  days  in  the  given  days,  is  times  the  initial  days. 

3.  Take  such  a  part  of  the  last  obtained  interest  as  will  equal 
the  greatest  aliquot  part  of  the  last  obtained  days  which  is  found  in 
the  remaining  days  of  the  given  time. 

4.  Continue  the  last  preceding  process  upon  each  following  result 
until  all  the  given  time  has  been  considered. 

VERIFICATION.  If  the  successive  aliquot  days  have  been  so  arranged 
that  the  terminating  aliquot  days  are  also  an  aliquot  of  the  initial  days,  by 
observing  if  the  last  obtained  aliquot  interest  is  also  derivable  from  the  initial 
interest,  it  will  not  only  be  a  verification  of  itself,  but  also  of  all  the  aliquots 
which  precede  it. 

NOTE  1.  If  the  given  time  is  expressed  in  years  and  days,  first  find  the 
interest  for  the  days,  and  verify.  Then  take  as  many  times  the  initial  in- 
terest (1  %  of  the  principal)  as  will  equal  the  rate  per  annum  multiplied  by 
the  number  of  years. 

NOTE  2.  If  the  given  interval  in  days  lacks  an  aliquot  of  the  initial  days  of 
being  1  more  time  the  initial  days,  it  will  usually  be  more  convenient  to  in- 
crease the  multiplier  of  the  initial  interest  by  1,  and  deduct  the  excess  of 
interest  from  the  resulting  product.  Thus,  to  find  the  interest  for  177  da.  at 
8  %,  first  find  the  interest  for  180  da.  (4  times  45  da.)  and  deduct  therefrom 
the  interest  for  3  da.  (-fa  of  180  da.). 

NOTE  3.  If  the  given  rate  is  not  an  exact  divisor  of  360  apply  234.  Or, 
find  the  interest  at  some  convenient  rate  which  is  a  divisor  of  360;  thus,  at 
3$  %,  if  more  convenient,  first  find  the  interest  at  3  %  and  add  i  of  itself;  or 
find  the  interest  at  4  %  and  deduct  i  of  itself. 


172  PERCENTAGE 

EXAMPLES  FOR  PRACTISE 

Verifying  each  result,  find  the  interest  of 

1.  $875.90  at  9  %  for  135  da.      5.  $297.50  at  3  %  for  165  da. 

2.  $462.75  at  4  %  for  210  da.      6.  $935.75  at  10  %  for  204  da. 

3.  $695.82  at  5  %  for  261  da.      7.  $392.18  at  12  %  for  316  da. 

4.  $715.60  at  8  %  for  318  da.      8.  $1586.35  at  4  J  %  for  205  da. 

Obtaining  the  time  by  the  30-day  method,  find  the  interest  of 
9.   $3272.24  from  July  20,  1902,  to  Feb.  17,  1908,  at  4  J  %. 

10.  $1873.16  from  Jan.  26,  1908,  to  Sept.  20,  1911,  at  10%. 

11.  $1384.20  from  Apr.  9,  1906,  to  Nov.  13,  1911,  at  3%. 

12.  $278.15  from  Nov.  30,  1905,  to  Aug.  10,  1908,  at  3J%. 

13.  $48.73  from  Apr.  28,  1906,  to  Nov.  26,  1910,  at  9%. 

Obtaining  the  exact  days,  find  the  interest  of 

14.  $435.76  from  Oct.  13,  1905,  to  July  30,  1911,  at  12%. 
16.   $2896.37  from  Aug.  19,  1897,  to  May  7,  1912,  at  5%. 

16.  $9183  from  June  16,  1909,  to  Mch.  24,  1912,  at  4  %. 

17.  $4238.75  from  May  31,  1904,  to  July  16,  1911,  at  4J  %. 

18.  $37.95  from  Sept.  24,  1903,  to  Dec.  13,  1910,  at  8%. 

INTERCHANGEABILITY  OF  PRINCIPAL  AND  TIME 

236.  INTRODUCTION,  (a)  In  232  it  was  seen  that  the  princi- 
pal and  the  time  in  days  are  uniform  factors  in  finding  the  inter- 
est at  any  rate  per  annum.  As  the  product  of  any  two  factors 
will  be  the  same  in  whatever  order  they  may  be  arranged,  it  must 
follow  that  the  product  of  the  time  in  days  multiplied  by  one- 
hundredth  of  the  principal  will  be  the  same  as  that  of  the  prin- 
cipal multiplied  by  one-hundredth  of  the  time  in  days;  that  is, 
the  principal  and  the  time  in  days  may  be  interchanged  at  the  conven- 
ience of  the  calculator  without  affecting  the  correctness  of  the  final 
result.  Thus,  the  interest  of  $180  for  240  da.  at  any  rate  per 
annum  will  be  the  same  as  the  interest  $240  for  180  da.  at  the 
same  rate.  Advantage  should  be  taken  of  this  fact  whenever 
it  is  seen  that  the  principal  regarded  as  days  can  be  separated 
into  fewer  aliquot  parts  of  the  initial  days  at  the  given  rate  than 
can  the  actual  time  in  days. 

(6)  This  interchange  of  principal  and  time  may  be  made  more 
frequently  advantageous  by  amplifying  the  principles  underlying 
the  processes  of  233  and  235.  Thus,  if  cutting  off  two  orders 


INTEREST  173 

from  the  right  of  the  dollars  in  any  principal  will  produce  the  in- 
terest of  that  principal  for  such  a  part  of  360  days  as  1  %  is  of  the 
given  rate  per  annum,  then  cutting  off  three  orders  from  the  right 
of  the  dollars  in  the  principal  (being  equivalent  to  taking  one- 
tenth  of  the  preceding  initial  interest)  must  produce  the  interest 
for  one-tenth  of  the  preceding  initial  days;  and  cutting  off  one 
order  from  the  right  of  the  dollars  in  the  principal  (being  equiva- 
lent to  taking  ten  times  the  same  initial  interest)  will  produce  the 
interest  for  ten  times  the  same  initial  days;  and  cutting  off  no 
order  from  the  right  of  the  dollars  in  the  principal  (being  equiva- 
lent to  taking  one  hundred  times  the  same  initial  interest)  will 
produce  the  interest  for  one  hundred  times  the  same  initial  days; 
etc.;  etc. 

(c)  Thus,  if  cutting  off  two  figures  from  the  right  of  the  dol- 
lars in  the  principal  will  produce  the  interest  of  that  principal 
at  6  %  for  60  da.,  then  cutting  off  three  figures  must  produce  the 
interest  for  one-tenth  of  60  da.,  or  6  da.;   cutting  off  one  figure, 
for  10  times  60  da.,  or  600  da.;  cutting  oft  no  figure,  for  100  times 
60  da.,  or  6000  da.     Similarly  at  9%  per  annum,  cutting  off  three 
orders  from  the  right  of  the  dollars  in  the  principal,  will  produce 
the  interest  for  4  da.;   cutting  off  two  orders,  for  40  da.;   1  order 
for  400  da.;    and  no  order,  for  4000  da.;   etc.;   etc. 

(d)  To  differentiate  these  processes,  cutting  off  three  orders 
is  suggestively  called  the  units'  rule  (3  units  of  days  at  12  %) ;  cut- 
ting off  two  orders,  the  tens'  rule  (3  tens  of  days) ;  cutting  off  one 
order,  the  hundreds'  rule  (3  hundreds  of  days);    and  cutting  off 
no  order,  the  thousands'  rule  (3  thousands  of  days). 

NOTE  1.  As  all  ordinary  interest  is  computed  upon  the  uniform  basis  of 
360  days  to  a  year,  if  the  given  time  is  expressed  in  years  and  days,  the  years 
should  be  reduced  to  days  upon  the  same  basis,  even  when  dates  are  given 
and  the  exact  number  of  days  is  found  in  the  fractional  part  of  a  year  in  the 
given  interval. 

NOTE  2.  If  the  given  principal  contains  cents,  no  advantage  will  be 
gained  by  interchanging,  for  the  cents  will  then  have  to  be  regarded  as  such 
a  fraction  of  a  day  as  they  are  of  one  dollar. 

ILLUSTRATIVE   EXAMPLES 

1.  Find  the  interest  of  $685.72  at  4%  for  27  days.  2.  Of 
$275.83  at  6  %  for  150  days.  3.  Of  $5000  at  8  %  for  123  days. 

SOLUTIONS 

(D  (2) 

|  $685.72  =  9  days'  int.  $27  |  5.83  =  600  days'  int. 

$2  I  057  16  =  27  "   "    $6  I  8  95  =  150  "   " 


174 


PERCENTAGE 

(3) 


$123 
J  of  $123  =  13 


=  4500  days'  int. 
666  =  500  " 


$136  I  67-=  5000 

EXAMPLES  FOR  PRACTISE 
Employing  the  fewest  figures,  find  the  interest  of 


1.  $496.75  at  6  %  for  24  da. 

2.  $715.25  at  6  %  for  54  da. 

3.  $563.70  at  6  %  for  18  da. 

4.  $758.62  at  6  %  for  27  da. 

5.  $435.80  at  6  %  for  150  da. 

6.  $847.35  at  6  %  for  200  da. 

7.  $6000  at  6  %  for  327  da. 

8.  $800  at  6  %  for  171  da. 

9.  $7500  at  6  %  for  192  da. 
10.  $400  at  6  %  for  259  da. 


11.  $400  at  8%  for  319  da. 

12.  $800  at  5%  for  153  da. 

13.  $4500  at  7%  for  137  da. 

14.  $300  at  9  %  for  273  da. 

15.  $600  at  4  J  %  for  169  da. 

16.  $2000  at  12  %  for  317  da. 

17.  $1800  at  10%  for  159  da. 

18.  $2500  at  6  %  for  3  yr.  253  da. 

19.  $7200  at  5  %  for  4  yr.  198  da. 

20.  $1800  at  3  %  for  2  yr.  147  da. 


ACCURATE  INTEREST 

237.  Accurate  interest  is  interest  computed  upon  the  accurate 
basis  of  365  days  to  a  common  year,  or  of  366  days  to  a  leap  year. 

NOTE  1.  All  disbursements  of  interest  by  the  governments  of  the  United 
States  and  of  foreign  nations,  all  calculations  of  interest  by  foreign  com- 
mercial houses,  and  many  computations  by  business  houses  in  the  United 
States,  especially  by  American  importers  of  foreign  goods,  are  made  upon  the 
accurate  basis. 

ILLUSTRATIVE  EXAMPLE 

Find  the  accurate  interest  of  $297.83  for  189  da.  at  8  %. 

SOLUTION  EXPLANATION.     First    find    the    in- 

$297.83  terest  of  the  given  principal  for  1  year 

.08  at  the  given  rate,   obtaining  $23.8264. 

As  the  accurate  interest  for  1  day  is  3^3- 
of  the  accurate  interest  for  1  year,  the 
accurate  interest  for  189  da.  must  be 
189  times  ^Jy,  or  ^f  f ,  of  the  accurate 
interest  for  one  year,  obtained  by  mul- 
tiplying $23.8264  by  the  numerator  (189) 
and  dividing  by  the  denominator  (365). 


23.8264 

189 

2144376 

1906112 

238264 


365)4503.1896($12.3375 


ACCURATE    INTEREST  175 

RULE.  Multiply  the  interest  for  one  year  at  the  given  rate  per 
annum  by  the  exact  number  of  days  in  the  given  fraction  of  a  year, 
and  divide  the  result  by  365  if  the  interval  is  part  of  a  common  interest 
year,  or  by  366  if  part  of  a  leap  interest  year. 

NOTE  2.  If  the  given  interval  is  more  than  one  year,  first  find  the  accurate 
interest  for  the  days  in  the  fraction  of  a  year,  and  then  find  the  interest  for  the 
years,  and  add  the  two  results. 

NOTE  3.  Always  obtain  the  accurate  time  [164,  or  Note  5,  233]  between 
two  dates  when  computing  accurate  interest. 

NOTE  4.  The  learner  is  cautioned  against  the  common  error  of  confound- 
ing the  calendar  year  with  the  interest  year.  A  calendar  year  commences  with 
Jan.  1  and  terminates  with  Dec.  31,  both  inclusive,  but  an  interest  year  com- 
mences with  the  date  on  which  the  debt  commences  to  draw  interest  and  ter- 
minates at  the  corresponding  date  of  the  next  following  year.  Thus,  an 
interest-bearing  debt  contracted  May  18,  1911,  and  paid  July  3,  1911,  has 
accrued  interest  for  46  days  of  the  366-day  year  commencing  May  18,  1911, 
and  terminating  May  18,  1912;  while  an  interest-bearing  debt  contracted 
May  18,  1912,  and  paid  July  3,  1912,  has  accrued  interest  for  46  days  of  the 
365-day  year  commencing  May  18,  1912,  and  terminating  May  18,  1913. 
The  leap  year  is  so  called  because  it  includes  the  leap  day  (Feb.  29);  therefore 
any  interest  year  which  includes  the  leap  day  must  be  a  leap  interest  year, 
though  the  leap  day  itself  may  not  be  included  in  the  given  interval  of  that 
leap  interest  year. 

EXAMPLES  FOR  PRACTISE 

Find  the  accurate  interest  of 

1.  $963.75  at  4  %  for  82  da.      7.  $296.75  at  7  %  for  93  da. 

2.  $725.62  at  3  %  for  174  da.     8.  $583.92  at  4£  %  for  335  da. 

3.  $286.40  at  5  %  for  230  da.     9.  $857.42  at  3  J  %  for  252  da. 

4.  $437.52  at  8  %  for  161  da.    10.  $728.19  at  5  %  for  3  yr.  87  da. 

5.  $582.37  at  6  %  for  271  da.    11.  $649.50  at  4£%  for  1  yr.  279  da. 

6.  $345.28  at  9  %  for  168  da.    12.  $484.32  at  4  %  for  4  yr.  176  da. 

13.  $2983.52  from  Aug.  18,  1903,  to  May  18,  1910,  at  9%. 

14.  $786.25  from  Mch.  14,  1907,  to  Nov.  28,  1907,  at  10%. 

15.  $685.97  from  Apr.  23,  1904,  to  Dec.  23,  1912,  at  8%. 

16.  $925.82  from  May  14,  1905,  to  Sept.  20,  1911,  at  12%. 

17.  $716.25  from  Jan.  3,  1909,  to  Feb.  18,  1912,  at  3i  %. 

18.  $1586.73  from  June  19,  1895,  to  Mch.  17,  1900,  at  7  %. 

19.  $6178.30  from  May  23,  1910,  to  Aug.  16,  1913,  at  5%. 

20.  $875.45  from  Oct.  12,  1909,  to  July  17,  1912,  at  3J  %. 

21.  $2495.18  from  Feb.  29,  1908,  to  Oct.  27,  1912,  at  4%. 


176  PERCENTAGE 

NOTE  5.  Reduce  the  lower  denominations  of  English  money  to  a 
decimal  of  a  £  as  in  111.  Ex.  2,  160,  compute  the  interest  in  the  usual  manner, 
and  reduce  the  decimal  part  of  a  £  in  the  obtained  answer  to  integers  of 
the  lower  denominations  as  in  111.  Ex.,  159. 

22.  Find  the  accurate  interest  of  £178  15s.  6d.  from  May  3, 
1908,  to  Nov.  18,  1913,  at  4%  per  annum. 

23.  What   is   the   accurate   interest    of   £3487  2s.  6d.  from 
Sept.  25,  1908,  to  June  10,  1913,  at  3%  per  annum? 

24.  What  is  the  accurate  interest  of  £625  18s.  3d.,  at  5%, 
from  Mch.21,  1910,  to  Sept.  6,  1913? 

25.  Find  the  accurate  interest  at  3|  %  per  annum  of  £345 
12s.  6d.  from  Nov.  3,  1905,  to  May  26,  1913. 

RELATION    OF    THE    FOUR    ELEMENTS 
OF    INTEREST 

238.  INTRODUCTION,  (a)  A  problem  in  interest  is  one  in 
which  three  of  the  four  elements  of  interest  are  given,  and  it  is 
required  to  find  the  omitted  fourth  element.  The  four  elements 
(principal,  rate  per  annum,  time,  interest  or  amount)  have  a  uni- 
form relation  to  each  other;  so  that  when  any  three  of  them  are 
given,  it  is  possible  to  complete  the  relation  by  finding  the  neces- 
sary omitted  element.  The  most  useful  methods  for  finding  the 
interest  when  it  is  the  omitted  element  have  already  been  pre- 
sented [231-236] .  It  now  remains  to  consider  problems  in  which 
the  interest  is  not  required  as  heretofore,  but  is  given,  together 
with  any  two  of  the  three  remaining  elements  (principal,  rate, 
time),  and  it  is  required  to  find  the  omitted  remaining  element. 
The  following  general  process  is  applicable  to  all  such  problems: 
(6)  Find  the  interest  or  the  amount  by  any  one  of  the 
preceding  methods  which  is  most  favored,  for  all  the  units  of  the 
given  elements,  and  computed  for  1  unit  of  the  required  element; 
then  divide  the  given  interest  or  amount  for  all  the  units  of  the 
required  element  by  the  obtained  interest  or  amount  for  1  unit 
of  the  required  element,  to  find  the  number  of  units  in  the 
required  element. 

SUMMARY.  1.  The  total  interest  or  amount  of  the  total  prin- 
cipal for  the  total  time  and  at  the  total  rate  %  per  annum,  is  a  product. 

2.  The  total  interest  or  amount  of  any  two  of  the  three  remaining 
elements  for  1  unit  of  the  required  element,  is  the  multiplicand. 

3.  The  number  of  units  in  the  required  element  is  the  multiplier. 


INTEREST  177 

PRINCIPAL,  INTEREST  AND  TIME  GIVEN,  TO  FIND  THE  RATE 

ILLUSTRATIVE  EXAMPLE 

239.   At  what  rate  per  annum  will  $758.36  produce  $13.65 
interest  in  216  days?  SOLUTION 

$7  I  58.36   =  int.  for    60  da..at  6  % 


22 

75  08  =    " 

"    180  " 

tt    tt 

3 

79  18  =    " 

"     30  " 

tt    tt 

75  83  =    " 

tt       6  tt 

tt    tt 

6)27 

30  09  =    " 

"  216  " 

it    ft 

4 

55  01  =    " 

"   216  " 

"  1% 

$4.5501)$13.6500  (  2,  practically  3  times  ) 
9  1002  I  the  int.  at  1  %,  or  3  %  ) 
4  5498  =  significant  remainder 

EXPLANATION.  First  find  the  interest  of  the  given  principal  ($758.36) 
for  the  given  time  (216  da.)  at  6  %,  obtaining  $27.3009  and  divide  by  6  to  find 
the  interest  at  1  %  ($4.5501). 

As  the  given  interest  ($13.65)  is  2  times  the  interest  at  1  %,  and  a  very 
significant  remainder  almost  equal  to  the  divisor  [206,  6],  it  should  be  regarded 
as  practically  3  times  the  interest  at  1  %,  or  3  %. 

SUMMARY.  The  required  rate  %  per  annum  will  be  as  many 
times  1  %,  as  the  given  interest  is  times  the  obtained  interest  at  1  % 
of  the  same  principal  for  the  same  time. 

NOTE  1.  If  the  remainder  from  the  final  division  is  insignificant,  follow 
205,  a;  if  significant,  follow  205,  b;  and  if  intermediate,  follow  205,  c.  In  prac- 
tise, the  division  is  rarely  exact. 

NOTE  2.  If  the  amount  [Note,  226]  is  given,  diminish  it  by  the  principal 
to  find  the  interest,  and  then  proceed  as  in  the  preceding  solution.  The  amount 
at  3  %  (103  %)  is  not  3  times  the  amount  at  1  %  (101  %). 

EXAMPLES  FOR  PRACTISE 
At  what  rate  %  per  annum  will 

1.  $596.85  produce  $19.70  interest  in  132  days? 

2.  $1237.29  produce  $43.99  interest  in  256  days? 

3.  $3278.45  produce  $163.92  interest  in  225  days? 

4.  $45.97  produce  62  cents  interest  in  108  days  [205,  c]? 

5.  $257.30  amount  to  $273.31  in  320  days? 

6.  $768.25  amount  to  $781.79  in  141  days  [205  c]? 


178  PERCENTAGE 

PRINCIPAL,  INTEREST  AND  RATE  GIVEN,  TO  FIND  THE  TIME 
ILLUSTRATIVE  EXAMPLE 

240.    In  what  time  will  $568.75  produce  $76.78  interest  at 
5  %  per  annum? 

SOLUTION  EXPLANATION.     The   in- 

$568.75  X  .05  =  $28.4375,  1  yr.'s  int.      terest  of  the  8iven  principal 


$28.4375)$76.7800(2  years  ($56875)  at  th;,spef  fic  rat,e 

V  per     annum     (5%)    equals 

$28.4375  for  1  year.     There- 

19.9050  fore  it  will  require  as  many 

years  to  produce  $76.78  in- 
terest  as  $28.4375  are  con- 


tained times  in  $76.78,  or  2 
5971  50  2  yr.  and  HBf  $  of  a  year. 

28.4375)7165.8000(25;  days  As   ordinary  interest  is 

^ftS7  f^n  computed  upon  the  basis  of 

360  da.  to  a  year,  there  will 

1478  300  be  as  many  days  in  if  f  f  ft 

1421  875  °f  a  yr->  as  it  contains  360ths. 

cfi  4.0  en  Hence,  reduce  the  fraction  of 

00  4ZOU  a  year  to  360ths  by  86,  ob- 

28  4375  taining  251  da.  and  a    very 

27  9875,  significant  significant  remainder  almost 

2  yr.  252  da.,  Ans.  ^  to  the  divisor  which 

should    be    counted    as    an- 

other day  [205,  &,]  making  the  required  time  2  yr.  252  da. 

SUMMARY.  The  required  time  will  be  as  many  times  one  year 
or  one  day,  as  the  given  interest  is  times  the  interest  for  one  year  or 
for  one  day}  of  the  same  principal  and  at  the  same  rate  per  annum. 

NOTE  1.  Apply  205,  a,  to  all  insignificant  remainders  from  the  final  divi- 
sion; and  205,  6,  to  all  significant  remainders. 

NOTE  2.  If  the  given  interest  is  less  than  the  interest  for  one  year,  divide 
it  at  once  by  the  interest  of  the  same  principal  for  1  day. 

EXAMPLES  FOR  PRACTISE 
In  what  time  will 

1.  $958.63  produce  $32.59  interest  at  8  %  per  annum? 

2.  $7285.15  produce  $166.65  interest  at  4|  %  per  annum? 

3.  $815.30  produce  $14.90  interest  at  7  %  ^per  annum? 

4.  $3825.75  produce  $130.50  interest  at  4%  per  annum? 

5.  $8270.75  amount  to  $10236.43  at  6  %  per  annum? 


INTEREST  179 

INTEREST,   TIME,   AND  RATE  GIVEN,   TO  FIND  THE 
PRINCIPAL 

ILLUSTRATIVE  EXAMPLE 

241.  What  principal  loaned  for  279  days  at  8%  per  annum, 
will  produce  $57.50  interest? 

SOLUTION 

2.79  -:-  45  =  .062,  int.  of  $1  at  8  %  for  279  da 
$57.50  4-  .062  =  927.42  times  $1,  or  $927.42. 

EXPLANATION.  8  %  per  annum  equals  8  cents  per  annum  for  each  dollar 
of  the  required  principal.  If  the  interest  of  $1  for  one  year  of  360  da.  is 

8  cents,  the  interest  of  $1  for  |  of  360  da.,  or  for  45  da.,  must  be  |  of  8  cents,  or 
1  cent.     If  the  interest  of  $1  is  1  cent  for  each  45  da.  of  the  given  time,  it  will 
be  as  many  cents  for  279  days  as  45  da.  are  contained  times  in  279  da.,  or 
6.2  cents.     Hence,  the  required  principal  must  be  as  many  times  $1,  as  the 
given  interest  of  the  required  principal  ($57.50)  is  times  the  interest  of  $1  of 
principal  ($.062),  for  the  same  time  and  at  the  same  rate  per  annum,  or  927.42 
times  $1. 

SUMMARY.  The  required  principal  will  be  as  many  times  $1, 
as  the  given  interest  of  the  required  principal  is  times  the  interest  of 
SI  for  the  same  time  and  at  the  same  rate  %  per  annum. 

NOTE.  To  find  the  interest  of  $1  for  any  given  time  and  at  any  given 
rate  per  annum,  divide  the  given  days  regarded  as  cents  by  such  a  part  of  360  as 
1  %  is  of  the  given  rate  per  annum.  Thus,  the  interest  of  $1  for  152  da.  at 

9  %  per  annum  equals  1.52  -f-  (360  -r-  9)  or  $.038  [Note,  227].     If  the  given 
rate  is  not  an  exact  divisor  of  360,  first  find  the  interest  of  $1,  at  some  other 
convenient  rate  which  is  an  exact  divisor  of  360,  and  add  to,  or  take  from,  the 
obtained  result  as  the  given  rate  is  greater  or  less  than  the  assumed  rate. 

EXAMPLES  FOR  PRACTISE 

What  principal  will  produce 

1.  $4.08  interest  at  3  %  in  84  days? 

2.  $295.58  interest  at  8  %  in  324  days? 

3.  $169.87  interest  at  6%  in  247  days? 

4.  $1524.75  interest  at  9  %  in  1  yr.  282  da.? 

5.  $5.58  interest  at  5  %  in  96  days? 

6.  $202.71  interest  at  7  %  in  144  days? 

7.  $445.73  interest  at  4%  in  2  yr.  315  da? 


180 


PERCENTAGE 


AMOUNT,  TIME,  AND  RATE  GIVEN  TO  FIND  THE  PRINCIPAL 

ILLUSTRATIVE  EXAMPLE 

242.   What  principal  will  amount  to  $349.59  in  51  days  at 
4  %  per  annum? 

FIRST   SOLUTION 

$.51  -T-  90  =  $.005f,  int.  of  $1  at  4% 
$1  +  $.005f  =  $1.005f,  amt.  of  $1  at  4  % 
$1. 005  f)  $349.59 
3  3 


3.017  )  1048.770(347.62  times  $1 
9051 
14367 
12068 
22990 
21119 
18710 
18102 

6080 
6034 


FIRST  SOLUTION.  The 
interest  of  $1  for  51  da. 
at  4  %  is  $.005|  [Note, 
241].  This,  added  to  $1, 
will  be  the  amount  of 
$1  for  the  same  time 
and  at  the  same  rate. 
As  the  given  amount 
($349.59)  is  347.62  times 
the  amount  of  $1  of 
principal  ($1.005|)  for 
the  same  time  and  at  the 
same  rate,  so  the  princi- 
pal which  produced  the 
given  amount  must  be 
347.62  times  $1,  or 
$347.62. 


SECOND   SOLUTION 

$349.59 
90 


$90.51)$31463.10(347.62  times  $1 
27153 
43101 
36204 

68970 
63357 
56130 
54306 
18240 
18102 

138 


SECOND  SOLUTION.  If  it  is 
seen  that  the  interest  of  $1  for  the 
same  time  and  at  the  same  rate 
[Note,  241]  will  include  an  in- 
convenient fraction,  as  it  usually 
does,  the  second  solution  will  be 
found  preferable. 

To  multiply  both  divisor  and 
dividend  by  the  same  number  will 
not  affect  the  value  of  the  quotient 
[Prin.  5,  34].  Therefore,  instead 
of  dividing  the  given  amount 
($349.59)  by  the  amount  of  $1  at 
the  same  rate  and  for  the  same 
time  ($1  +  '•&)  as  in  the  first 
solution,  it  will  usually  involve  less 
labor  to  divide  90  times  the  given 
amount  ($349.59  X  90  =  $31463.10) 


COMPOUND    INTEREST  181 

by  90  times  the  amount  of  $1  for  the  given  time  and  at  the  given  rate 
($1  X  90  +  •$&  X  90  =  90.51).  Hence,  multiply  the  given  amount  by  such 
a  part  of  860  as  1  %  is  of  the  given  rate  per  annum  (in  the  111.  Ex.,  by  360  -j- 4, 
or  90);  and  divide  the  result  by  the  same  quotient  (90)  increased  by  the  days 
regarded  as  hundredths  (.51),  or  by  90.51. 

SUMMARY.  The  required  principal  is  as  many  times  $1  of  prin- 
cipal, as  the  given  amount  of  the  required  principal  is  times  the  ob- 
tained amount  of  $1  of  principal  for  the  same  time  and  at  the  same 
rate  %  per  annum. 

NOTE.  If,  in  using  the  second  solution,  the  given  rate  (as  7  %,  3|  %,  etc.) 
is  not  an  exact  divisor  of  360,  the  resulting  fraction  may  be  avoided  by  mul- 
tiplying the  given  amount  by  360,  and  dividing  the  result  by  360  increased  by  as 
many  times  the  days  of  prepayment  regarded  as  hundredths  as  the  given  rate  is 
times  1  %. 

Thus,  in  the  111.  Ex.,  ($349.59  X  360)  -:-  (360  +  .5T~X4)  =  $347.62. 
That  is,  instead  of  dividing  90  times  the  given  amount  by  90  times  the  amount 
of  $1,  as  in  the  second  solution,  by  this  process  4  times  90  times,  or  360  times, 
the  given  amount  is  divided  by  4  times  90  times  the  amount  of  $1  (90  X  4  -f- 
.51  X  4)  or  by  362.04  [Prin.  5,  34]. 

EXAMPLES  FOR  PRACTISE 

What  principal  will  amount  to 

1.  $628.73  in  243  days  at  8%  per  annum? 

2.  $1965.37  in  162  days  at  6%  per  annum? 

3.  $748.26  in  117  days  at  3%  per  annum? 

4.  $3562.80  in  327  days  at  7  %  per  annum  [See  Note]? 
6.   $523.42  in  1  yr.  257  da.  at  4  J  %  per  annum? 

6.  $2685.72  in  2  yr.  168  da.  at  3  J  %  per  annum? 

7.  What  principal  will  amount  to  $986.35  on  May  16,  1911,  if 
loaned  on  Aug.  28,  1910,  at  5%  (exact  days)? 

8.  What  sum  of  money  loaned  Apr.   16,   1911  at  4%,  will 
amount  to  $1628.15  on  Nov.  27,  1911  (time  by  30-da.  method)? 


COMPOUND  INTEREST 

243.  Compound  interest  is  interest  computed  in  the  usual 
manner,  but  which,  at  the  end  of  a  stipulated  period  of  time,  is 
compounded  with  (that  is,  added  to)  the  principal  for  that  period, 
to  form  a  new  principal  for  the  next  following  period. 


182  PERCENTAGE 

ILLUSTRATIVE  EXAMPLE 

Find  the  interest  of  $678.45  for  1  yr.  320  da.,  if  compounded 
semi-annually  at  6%  per  annum. 

SOLUTION  EXPLANATION.     At  6  % 

»^TO  A  r       AC  •      •      i\  Per  annum>  the  correspond- 

$678.45      (first  principal)  ing  rate  is  i  of  6  %>  or 

20.3535  (3%  Of  $678.45)  3  %,  for  one-half  a  year. 

698.8035  (first  amount)  l    vr-  =  2    half-years; 

20  QfU1   (1  07  nf  ftfiQK  SO*}^  and  32°  da>  =  *  half  year 

20-9b41  ('  (i  of   360  da.  =  180  da.) 

719.7676  (second  amount)  and  140  da.  of  a  fourth 

21. 5930  (3%  Of  $719.7676)  half  year.  Hence,  1  yr. 

741.3606  (third  amount)  ^20  f  =  3  half-years  and 

17.2984  (int.  of  $741.36  for  140  da.)  ^secure  accuracy  in 

758.6590  (final  amount)  the   final  result,   carry  all 

678.45  (original  principal)  intermediate  results  to  four 

decimal  places.  As  hun- 

80.21-  (comp.  int.  for  1  yr.  320  da.)  dredths  x  hundredths  = 

ten-thousandths,  so  to  ob- 
tain the  necessary  four  decimal  places,  multiply  the  cents'  order  of  the  principal 
(hundredths)  by  the  rate  %  per  period  (3  hundredths),  and  write  the  first 
order  of  the  product  underneath  the  principal  but  two  places  to  the  right 
of  the  cents'  order  of  the  principal  (that  is,  in  the  ten-thousandths'  order). 
Add  the  obtained  interest  for  the  first  period  ($20.3535)  to  the  first  prin- 
cipal ($678.45),  to  find  the  first  amount  ($698.8035),  which  will  constitute  a 
new  principal  for  the  second  half-yearly  interest  period. 

Continue  this  process  as  many  times  as  there  are  complete  interest  periods 
(3  times).  Then  find  the  interest  of  the  amount  for  the  last  complete  period 
($741.3606)  for  the  fraction  of  the  fourth  period  (140  da.)  at  6  %  per  annum 
by  233,  obtaining  $17.2984,  which  added  to  the  last  obtained  amount 
($741.3606)  will  produce  the  final  amount  ($758.66  -). 

The  difference  between  the  original  principal  ($678.45)  and  the  final 
amount  ($758.66)  will  be  the  required  compound  interest  ($80.21  — ). 

NOTE  1.  If  the  given  time  contains  a  fraction  of  a  year  expressed  in  days, 
regard  180  days  thereof  (if  it  contains  so  many)  as  another  half-year  if  the 
compounding  of  interest  is  semi-annual;  or  regard  each  90  days  thereof  as 
another  quarter,  if  the  compounding  of  interest  is  quarterly.  If  dates  are 
given,  and  there  is  a  fraction  of  a  year,  regard  each  successive  six  calendar 
months  thereof,  irrespective  of  the  number  of  days  they  may  contain,  as 
another  half-year  if  the  compounding  is  semi-annual,  or  each  three  calendar 
months  thereof  as  another  quarter  if  the  compounding  is  quarterly;  and 
compute  the  time  for  any  possible  remaining  fractional  part  of  an  interest 
period  in  the  customary  manner. 


COMPOUND   INTEREST  183 

EXAMPLES  FOR  PRACTISE 

Find  the  compound  interest  of 

1.  $4375.80  for  3  yr.  at  6  %  per  annum,  comp.  annually. 

2.  $965.25  for  4  yr.  at  4  %  per  annum,  comp.  half-yearly. 

3.  $1825.30  for  2  yr.  at  5  %  per  annum,  comp.  quarterly. 

4.  $2469.75  for  4  yr.  at  7  %  per  annum,  comp.  annually. 

5.  $6138.26  for  2  yr.  at  8  %  per  annum,  comp.  quarterly. 

6.  $594.65  for  1  yr.  270  da.  at  6  %,  comp.  half-yearly. 

7.  $763.40  for  1  yr.  195  da.  at  4  %  per  annum,  comp.  quarterly. 

8.  $298.84  for  2  yr.  78  da.  at  3  %,  comp.  half-yearly. 

9.  $3297  for  1  yr.  54  da.  at  5%  per  annum,  comp.  quarterly. 

10.  $682  for  2  yr.  173  da.  at  4^  %  per  annum,  comp.  annually. 

11.  $456.75  for  2  yr.  135  da.  at  9  %,  comp.  half-yearly. 

12.  $791.62  for  1  yr.  198  da.  at  8%,  comp.  quarterly. 

13.  No  payment  of  interest  having  been  previously  made, 
what  is  the  amount  due  Oct.  23,  1911,  upon  a  debt  of  $578.60, 
contracted  May  12,  1908,  at  5  %  per  annum,  interest  to  be  com- 
pounded annually  if  not  paid  (time  by  compound  subtraction)? 

14.  What  payment  on  June  17,  1912,  will  discharge  a  debt 
of  $1837.15,  incurred  Nov.  25,   1910,  drawing  interest  at  4%, 
payable  quarterly,  if  the  debtor  agreed  to  compound  all  deferred 
interest  payments,  and   had   permitted  the  second  and  fourth 
interest  payments  to  lapse  (time  in  exact  days)? 

15.  All  the  interest  payments  having  lapsed,  what  was  the 
balance  due  Sept.  10,  1910,  upon  a  note  of  $725.65,  dated  June  21, 
1908,  drawing  interest  at  4  J  %  per  annum,  payable  semi-annually 
and  specifying  that  the  interest  is  to  be  compounded  if  not  paid 
(time  by  comp.  subtraction)? 

16.  What  was  the  amount  due  Aug.  17,  1912,  upon  a  note  of 
$648.75,  dated  Feb.  17,  1911,  drawing  interest  from  date  at  5% 
per  annum,  and  to  be  compounded  quarterly  if  not  paid,  if  no 
payment  had  been  made  previous  to  final  settlement  (time  by 
compound  subtraction)? 

NOTE  2.  Much  labor  in  computing  compound  interest  may  be  saved  by 
the  use  of  a  complete  compound  amount  table.  The  following  abridgment 
will  sufficiently  illustrate  its  use: 


184 


PERCENTAGE 

COMPOUND   AMOUNT  TABLE 


Periods  | 

1% 

2% 

3% 

4% 

5% 

6% 

Periods  || 

1 

1.0100  000 

1.0200  0000 

1.0300  0000 

1.0400  0000 

1.0500  000 

1.0600  000 

1 

2 

1.0201  000 

1.0404  0000 

1.0609  0000 

1.0816  0000 

1.1025  000 

1.1236  000" 

2 

3 

1.0303  010 

1.0612  0800 

1.0927  2700 

1.1248  6400 

1.1576  250 

1.1910  160 

3 

4 

1.0406  040 

1.0824  3216 

1.1255  0881 

1.1698  5856 

1.2155  063 

1.2624  770 

4 

5 

1.0510  101 

1.1040  8080 

1.1592  7407 

1.2166  5290 

1.2762  816 

1.3382  256 

5 

6 

1.0615  202 

1.1261  6242 

1.1940  5230 

1.2653  1902 

1.3400  956 

1.4185  191 

6 

7 

1.0721  354 

1.1486  8567 

1.2298  7387 

1.3159  3178 

1.4071  004 

1.5036  303 

7 

8 

1.0828  567 

1.1716  5938 

1.2667  7008 

1.3685  6905 

1.4774  554 

1.5938  481 

8 

9 

1.0936  853 

1.1950  9257 

1.3047  7318 

1.4233  1181 

1.5513  282 

1.6894  790 

9 

10 

1.1046  221 

1.2189  9442 

1.3439  1638 

1.4802  4428 

1.6288  946 

1.7908  477 

10 

11 

1.1156  683 

1.2433  7431 

1.3842  3387 

1.5394  5406 

1.7103  394 

1.8982  986 

11 

12 

1.1268  250 

1.2682  4179 

1.4257  6089 

1.6010  3222 

1.7958  563 

2.0121  965 

12 

13 

1.1380  933 

1.2936  0663 

1.4685  3371 

1.6650  7351 

1.8856  491 

2.1329  283 

13 

14 

1.1494  742 

1.3194  7876 

1.5125  8972 

1.7316  7645 

1.9799  316 

2.2609  040 

14 

15 

1.1609  690 

1.3458  6834 

1.5579  6742 

1.8009  4351 

2.0789  282 

2.3965  582 

15 

16 

1.1725  786 

1.3727  8570 

1.6047  0644 

1.8729  8125 

2.1828  746 

2.5403  517 

16 

17 

1.1843  044 

1.4002  4142 

1.6528  4763 

1.9479  0050 

2.2920  183 

2.6927  728 

17 

18 

1.1961  475 

1.4282  4625 

1.7024  3306 

2.0258  1652 

2.4066  192 

2.8543  392 

18 

19 

1.2081  090 

1.4568  1117 

1.7535  0605 

2.1068  4918 

2.5269  502 

3.0255  995 

19 

20 

1.2201  900 

1.4859  4740 

1.8061  1123 

2.1911  2314 

2.6532  977 

3.2071  355 

20 

DIRECTIONS.  1.  To  find  the  compound  amount,  multiply  the  amount 
of  $1  for  the  given  number  of  periods  and  at  the  rate  %  per  period,  as  given 
in  the  above  table,  by  the  given  principal. 

2.  To  find  the  compound  interest,  deduct  $1  from  the  amount  of  $1  for 
the  given  number  of  periods  and  at  the  rate  %  per  period,  as  given  in  the 
above  table,  to  find  the  compound  interest  of  $1;  and  multiply  by  the  given 
principal. 

CAUTION.  The  learner  is  warned  against  the  error  of  supposing  that  the 
compound  amount  or  compound  interest  at  one  rate  per  period  is  proportioned 
to  the  compound  amount  or  compound  interest  at  some  other  rate  for  the  same 
time;  that  is,  that  the  compound  amount  or  compound  interest  at  7  %  is  one- 
sixth  more  than  the  compound  amount  or  compound  interest  at  6  %.  For, 
$1  +  the  interest  of  that  $1,  at  7  %,  cannot  evidently  be  one-sixth  more  than 
another  $1  +  the  interest  of  that  other  $1,  at  6  %;  that  is,  that  ($1  X  1.07  X 
1.07  X  1.07)  —  $1,  or  the  compound  interest  of  $1  at  7  %,  for  3  years,  can  be 
one-sixth  more  than  ($1  X  1.06  X  1.06  X  1.06)  -  $1,  or  the  compound 
interest  of  $1  at  6  %  for  the  same  number  of  years.  If  their  multiplicands 


TRUE   DISCOUNT  185 

are  the  same,  simple  products  will  be  proportioned  to  their  respective  multi- 
pliers; but  continued  products  have  constantly  varying  multiplicands,  and 
cannot  therefore  be  thus  proportioned. 

17.  Find  the  compound  amount  of  $785.30  for  8  years  at  6  % 
per  annum  compounded  semi-annually. 

SOLUTION 
8  years  =  16  half-years;  6  %  per  annum  =  3  %  per  half-year. 

$1.60470644  =  amt.  of  $1  for  16  periods  at  3%  per  period. 

X  785.30  =  given  number  of  dollars. 
$1260.175967332  =  comp.  amt.  of  $785.30  for  16  periods  at  3%. 

Find  from  the  table  the  compound 

18.  Amount  of  $968.35  for  13  yr.  at  6  %  per  yr.,  comp.  yearly. 

19.  Amount  of  $1873.42  for  9  yr.  at  5  %  per  yr.,  comp.  yearly. 

20.  Amount  of  $628.95  for  6  yr.  at  8  %,  comp.  half-yearly. 

21.  Amount  of  $716.25  for  8  yr.  at  4%,  comp.  half-yearly. 

22.  Interest  of  $845.73  for  3  yr.  at  12  %,  comp.  quarterly. 

23.  Interest  of  $467.28  for  4  yr.  at  4  %,  comp.  quarterly. 

24.  Interest  of  $2875.50  for  7  yr.  at  6  %,  comp.  yearly. 

25.  Interest  of  $943.25  for  5  yr.  at  10%,  comp.  half-yearly. 


TRUE  DISCOUNT 

244.  INTRODUCTION.  The  sum  to  be  paid  in  discharge  of  a 
debt  will  vary  with  the  time  of  payment.  If  paid  on  the  due  date, 
the  sum  to  be  paid  will  be  the  debt  itself,  without  any  increase  for 
interest  or  decrease  for  discount;  if  paid  after  it  is  due,  the  sum 
to  be  paid  will  be  the  original  debt  increased  by  the  accrued  inter- 
est thereon  from  its  due  date  to  the  date  of  postpayment;  and  if 
paid  before  it  is  due,  the  sum  to  be  paid  will  be  the  original  debt 
diminished  by  the  interest  on  the  prepayment  from  the  date  of 
prepayment  to  the  due  date  of  the  debt.  That  is,  interest  is  a 
compensation  for  the  use  of  money  [224,  b] .  Therefore,  if  money 
is  used  after  the  stipulated  time  of  payment,  the  debtor  should 
pay  compensatory  interest  thereon  for  every  day  during  which 
the  creditor  has  been  deprived  of  the  use  of  what  was  lawfully 
his;  and  if  money  is  paid  before  the  stipulated  time,  the  debtor 
should  receive  compensatory  interest  thereon  for  every  day  dur- 
ing which  he  has  deprived  himself  of  the  use  of  that  which  was 


186  PERCENTAGE 

legally  his.  In  either  case,  the  question  is,  what  sum  paid  at  a 
preceding  or  subsequent  date,  is  equivalent  to  a  given  sum  which 
is  legally  payable  at  a  certain  fixed  date;  or,  with  particular  ref- 
erence to  a  prepayment  as  in  true  discount,  what  sum  of  money 
placed  at  interest  on  the  date  of  prepayment  at  the  current  rate  per 
annum  will  amount  to  the  same  sum  as  the  given  debt  on  the  date 
upon  which  that  debt  is  due. 

245.  The  worth  of  a  debt  is  its  value  as  expressed  in  the 
legal  currency  of  the  country  in  which  it  is  estimated. 

246.  The  present  worth  of  a  debt  is  its  value  on  the  day  it 
is  paid. 

NOTE.  The  true  or  correct  value  of  a  debt  due  at  a  certain  future  time 
and  paid  before  its  maturity,  that  is,  the  present  worth  of  that  debt  by  true 
discount,  must  be  such  a  principal  [226]  as  loaned  on  the  day  of  prepayment 
at  the  current  rate  per  annum,  will  amount  to  that  debt  on  the  day  of  maturity 
[242];  for  the  creditor  will  then  have  received  the  principal  and  the  use  (in- 
terest) of  that  principal  for  the  days  of  prepayment,  and  if  the  two  combined 
(amount)  are  equal  to  the  debt  at  maturity,  it  must  be  an  equitable  discharge 
of  that  debt. 

247.  The  future  worth  of  a  debt  is  its  value  on  the  day  it 
is  due. 

NOTE.  Future  refers  to  any  time  yet  to  come;  but  as  here  used,  it  is 
restricted  to  that  specific  future  time  at  which  a  debt  falls  due.  Hence,  the 
future  worth  of  a  non-interest  bearing  debt  must  be  the  debt  itself;  but  the 
future  worth  of  an  interest  bearing  debt  must  be  the  debt  plus  the  accrued 
interest  thereon  from  the  date  upon  which  it  commenced  to  draw  interest  to 
its  due  date. 

248.  True  discount  is  the  deduction  from  the  future  worth  of 
a  debt  which  is  equal  to  the  interest  upon  its  present  worth  for 
the  days  of  prepayment. 

NOTE.  True  discount  is  so-called  because  it  is  the  true  or  correct  deduc- 
tion; that  is,  it  is  the  compensatory  interest  upon  the  exact  sum  of  the  use  of 
which  the  debtor  has  deprived  himself  by  paying  the  debt  before  it  was  legally 
due. 

ILLUSTRATIVE  EXAMPLE 

249.  A  debt  of  $764.50,  legally  due  Oct.  19,  1911,  was  paid 
June  25,  1911.     What  was  the  sum  paid,  allowing  true  discount 
at  6%?     What  was  the  true  discount? 


TRUE    DISCOUNT  187 

SOLUTION  EXPLANATION.    The  present 

solution  is  similar  to  the  second 

June  25  to  Oct.  19  =  116  da.  solution  of  111.  Ex.,  242,  to  which 

60.       $764.50  the  student  is  referred  for  a  full 

1.16  60  explanation. 

6l7l6)45870.00($750  Present  Worth.  Divide  60  times  the  per- 

428 1 2  centage  of  future  worth  ($764 . 50 

X  60  =  $45870)  by  60  times  its 
specific  rate  (1  +  -Vfc6)  X  60, 

30580  or  61.16,  obtaining  $750  as  the 

Q  base,  or  required  present  worth. 

If  preferable,  the  first  solu- 
$764.50  -  $750.  =  $14.50,  true  disct.   tion  of  111.  Ex.,  242,  may  be 

used. 

VERIFICATION.  Find  the  interest  of  the  obtained  present  worth  ($750) 
for  the  days  of  prepayment  (116),  and  add  this  interest  ($14.50)  to  the 
present  worth  ($750).  If  the  result  ($764.50)  equals  the  future  worth,  the 
solution  may  be  regarded  as  mechanically  correct. 

250.  SUMMARY.  True  discount  is  an  application  of  242,  the 
present  worth  of  a  debt  corresponding  to  the  principal;  the  true 
discount,  to  the  interest  of  the  principal  for  the  days  of  prepay- 
ment; the  future  worth,  to  the  amount;  the  days  of  prepayment 
to  the  time;  and  the  rate  %  of  discount,  to  the  rate.  Hence,  to 
find  the  present  worth: 

RULE.  Multiply  the  future  worth  by  such  a  part  of  860  as  1  % 
is  part  of  the  given  rate  per  annum,  and  divide  the  product  by  the  same 
part  of  360  increased  by  the  days  of  prepayment  regarded  as  hun- 
dredths. 

NOTE  1.  To  avoid  fractions,  if  the  rate  per  annum  of  discount  is  not  an 
aliquot  part  of  360,  multiply  the  future  worth  by  360,  and  divide  the  resulting 
product  by  360  increased  by  as  many  times  the  days  of  prepayment  regarded 
as  hundredths,  as  the  given  rate  is  times  1  %.  Thus,  in  the  111.  Ex.,  ($764.50 
X  360)  -5-  (360  +  1.16  X  6)  =  $750  [Note,  242]. 

NOTE  2.  To  find  the  present  worth  of  an  interest-bearing  debt,  first  find 
its  future  worth  (amount  due  at  maturity,  Note,  247) ;  and  then  proceed  as 
with  other  future  worths. 

NOTE  3.  If  the  time  of  prepayment  is  expressed  in  years  or  months, 
reduce  to  days  upon  the  basis  of  360  days  to  each  year,  or  of  30  days  to  each 
month.  If  dates  are  given,  find  the  days  of  prepayment  in  accordance  with 
local  usage. 


188  PERCENTAGE 

EXAMPLES  FOR  PRACTISE 
Find  the  present  worth  and  the  true  discount  of 

1.  $968.75  at  6  %,  if  paid  49  days  before  maturity. 

2.  $723.46  at  5  %,  if  paid  127  days  before  maturity. 

3.  $2965.80  at  4 }  %,  if  paid  213  days  before  maturity. 

4.  $7816.35  at  8%,  if  paid  53  days  before  maturity. 

6.   $527.90  at  7%,  if  paid  165  da.  before  maturity  [Note  1]. 

6.  $845.12  at  3?  %,  if  paid  87  days  before  maturity. 

7.  $693.50  at  3  %,  paid  2  yr.  139  da.  before  maturity  [Note  3]. 

8.  $378.65  at  4%,  paid  1  yr.  173  da.  before  maturity. 

9.  Find  the  true  discount  on  $1945.80  at  5  %,  if  paid  98  days 
before  it  is  due. 

10.  What  is  the  true  discount  on  a  debt  of  $635.70,  at  6  %,  if 
paid  253  days  before  maturity? 

11.  A  house  was  bought  for  $3600,  payable  one-half  cash, 
and    the    remainder    in    6    months    without    interest.      What 
equivalent  single  cash  payment  can  be  made  on  date  of  purchase, 
allowing  true  discount  on   the  deferred  payment   at   5%  per 
annum? 

12.  The  same  kind  of  broadcloth  can  be  bought  from  A  at 
$3.50  per  yard  and  2  months'  credit,  and  from  B  at  $3.58  per 
yd.  and  6  months'  credit.     Of  whom  will  it  be  more  advantageous 
to  buy;  and  how  much  will  be  gained  by  purchasing  95  yards 
at  the  more  favorable  of  these  terms,  allowing  true  discount  at 
6  %  per  annum? 

13.  Merchandise  amounting  to  $726.85  was  bought  on  May 
23,  1911,  at  90  days,  upon  which  a  trade  discount  of  25%,  20%, 
and  10  %  was  allowed.     What  cash  payment  on  July  7,  1911,  will 
be  in  full  payment  of  the  purchase,  allowing  true  discount  at 
4J  %  per  annum,  and  computing  the  exact  time  in  days? 

14.  The  same  brand  of  flour  can  be  bought  from  one  dealer 
at  $6.25,  and  4  months'  credit,  and  from  another  dealer  at  $6.20 
cash.     How  much  per  barrel  will  be  saved  by  purchasing  at  the 
more  advantageous  of  these  terms,  allowing  true  discount  at  5  % 
per  annum? 

15.  A  debt  of  $1785.90,  bearing  interest  at  5%  from  Apr. 
17,  1911,  and  legally  due  Nov.  13,  1911,  was  paid  Aug.  6,  1911. 


BANK   DISCOUNT  189 

What  was  the  sum  paid,  allowing  true  discount  at  6  %,  and  com- 
puting the  exact  time  in  days? 

16.  Merchandise  amounting  to  $948.60  was  bought  on  3 
months'  credit.  What  payment  made  1  mo.  12  da.  after  the 
date  of  purchase  will  discharge  the  debt,  allowing  true  discount 
at  4  %? 

BANK  DISCOUNT 

251.  Bank  discount  is  a  deduction  from  the  future  worth  of 
a  debt  which  is  equal  to  the  interest  upon  that  future  worth  for 
the  number  of  days  it  is  paid  before  maturity. 

NOTE.  As  true  discount  is  the  difference  between  the  present  worth  of  a 
debt  (the  principal  in  true  discount,  248)  and  its  future  worth  (the  principal 
in  bank  discount),  so  the  difference  between  the  two  discounts  must  be  the 
interest  upon  the  difference  between  the  present  worth  and  the  future  worth, 
that  is,  the  interest  upon  the  true  discount. 

252.  The  maturity  of  a  debt  is  the  date  upon  which  it  is 
legally  due. 

NOTE  1.  By  the  law  of  a  few  States,  if  a  debt  is  expressed  in  the  form  of 
a  note  or  draft,  3  days,  called  days  of  grace,  are  added  to  the  nominal  time 
(time  named  upon  the  face  of  a  note  or  draft)  in  determining  its  maturity. 
The  nominal  maturity  of  a  note  or  draft  is  therefore  the  expiration  of  the  time 
named  upon  its  face;  and  the  legal  maturity,  if  grace  is  allowed,  will  be  3 
days  thereafter.  If  no  grace  is  allowed,  as  in  most  States,  the  two  maturities 
fall  upon  the  same  date. 

NOTE  2.  In  finding  the  maturity  of  commercial  paper  which  is  drawn 
payable  a  given  number  of  months  after  date,  if  it  should  fall  nominally  due 
on  the  29th,  30th,  or  31st  of  a  month  which  contains  fewer  than  that  number 
of  days,  it  will  be  nominally  due  on  the  last  day  of  such  a  month;  and  if  grace 
is  allowed,  it  will  be  legally  due  on  the  third  day  of  the  next  following  month 
[Note  2, 163]. 

253.  The  term  of  discount  is  the  number  of  days  between  the 
date  of  discount  and  the  date  of  legal  maturity. 

NOTE.  In  computing  the  term  of  discount,  the  custom  of  many  cities  is 
to  include  within  the  interval  both  the  day  of  discount  and  the  day  of  maturity, 
which  is  equivalent  to  calculating  discount  for  1  day  more  than  the  time  as 
ordinarily  computed. 

254.  The  proceeds  of  a  note  or  draft  is  its  cash  value  on  the 
day  upon  which  it  is  discounted. 


190  PERCENTAGE 

NOTE.  The  proceeds  is  what  remains  of  the  future  worth  after  deducting 
the  interest  thereon  for  the  days  of  prepayment. 

255.  Identification  of  terms.  1.  The  sum  due  at  maturity  is 
the  principal  (multiplicand).  2.  The  specific  rate  of  discount  or 
of  proceeds  is  the  rate  (multiplier).  3.  The  discount  or  the  proceeds 
is  the  percentage  (product). 

NOTE.  As  the  proceeds  of  a  debt  equals  the  sum  due  at  maturity  minus 
the  interest  thereon  for  the  days  of  prepayment  [Note,  264],  so  the  specific 
rate  of  proceeds  must  equal  the  rate  of  the  sum  due  at  maturity  or  base  (100  %) 
minus  the  specific  rate  of  interest  for  the  days  of  prepayment  at  the  given  rate 
of  interest  per  annum  [Note,  227]. 


TO  FIND  THE  BANK  DISCOUNT  OR  THE  PROCEEDS 
ILLUSTRATIVE    EXAMPLE 

256.  A  note  of  $875.90,  dated  May  24,  1912,  legally  payable 
90  days  after  date,  was  discounted  on  July  5,  1912,  at  6  %.  Find 
the  date  of  maturity,  the  term  of  discount,  the  discount,  and  the 
proceeds. 

SOLUTION 

May  24  +  90  days  =  Aug.  22,  maturity. 
July  5  to  Aug.  22  =  48  da.,  term  of  discount. 
$875.90  =  discount  for  6  da.  at  6  %. 


007  2    =  discount  for  48  da.  at  6  %. 
$875.90  -  $7.01  =  $868.89,  proceeds 

EXPLANATION.  90  days  after  the  date  of  the  note  (May  24)  will  require 
the  7  remaining  days  of  May  +  the  30  da.  of  June  +  the  31  da.  of  July  ( =  68 
da.),  and  22  da.  of  Aug.,  thus  making  Aug.  22,  the  expiration  of  the  90th  day, 
or  the  date  of  maturity. 

From  July  5  (the  date  of  discount)  to  Aug.  22  inclusive  (the  date  of 
maturity)  =  48  da.  (the  days  of  prepayment,  or  term  of  discount). 

Interest  of  the  sum  due  at  maturity  (face  of  note,  $875.90)  for  the  time 
of  prepayment  (48  da.)  equals  the  compensatory  discount  for  paying  the  note 
before  it  is  legally  due  ($7.01  -). 

The  worth  at  maturity  ($875.90)  minus  the  discount  for  prepayment 
($7.01)  equals  the  proceeds  or  present  worth  of  the  note  on  the  date  of  dis- 
count ($868.89). 


BANK    DISCOUNT 


191 


SUMMARY.  1.  Bank  Discount  is  the  interest  upon  the  face* 
of  a  non-interest-bearing  note,  or  upon  the  amount  due  at  maturity 
of  an  interest-bearing  note,  or  upon  the  future  worth  of  any  form  of 
debt,  for  the  number  of  days  it  is  paid  before  it  is  legally  due.  2. 
The  proceeds  equal  the  sum  due  at  maturity  minus  the  bank  discount. 

*NOTE  1.  The  face  value,  or  as  it  is  commonly  abbreviated,  the  face  of  a 
note,  is  the  value  expressed  in  the  body  of  the  note,  exclusive  of  interest. 

NOTE  2.  An  interest-bearing  note  has  three  values:  its  face  value,  its 
value  at  maturity  (future  worth),  and  its  cash  value  on  the  day  it  is  discounted 
or  paid  (proceeds) .  In  non-interest-bearing  notes,  the  face  value  also  expresses 
the  value  at  maturity  or  future  worth. 

NOTE  3.  In  allowing  interest  upon  interest-bearing  notes,  when  the  time 
of  payment  is  expressed  in  months,  most  banks  compute  only  30  days  to  each 
of  the  expressed  months;  but,  in  computing  the  term  of  discount  upon  such 
notes,  the  exact  number  of  days  of  prepayment  is  found. 


EXAMPLES  FOR  PRACTISE 

[To  THE  STUDENT.  In  the  following  examples,  allow  no  grace  and  find 
the  exact  number  of  days  in  the  term  of  discount,  unless  otherwise  directed 
by  the  Teacher  to  conform  with  a  differing  local  usage  such  as  Note  1,  252 
or  Note,  263.] 

Find  date  of  maturity,  term  of  disc.,  disc.,  and  proceeds. 


Face  of 

Date  of 

Term  of 

Date  of 

%of 

note 

note 

note 

disc. 

disc. 

1. 

$596.75 

May  17,  1911 

90  days 

June  12,  1911 

5   % 

2. 

$842.90 

Nov.  25,  1911 

4  months 

Jan.  7,  1912 

6   % 

3. 

$285.32 

Feb.  15,  1912 

60  days 

Mch.  9,  1912 

4J% 

4. 

$1763.25 

Dec.  3,  1911 

3  months 

Feb.  18,  1912 

7   % 

6. 

$934.65 

Oct.  19,  1910 

30  days 

Oct.  23,  1910 

4   % 

6. 

$2481.12 

Mch.  31,  1912 

3  months 

Apr.  20,  1912 

3   % 

7. 

$672.85 

Sept.  24,  1911 

90  days 

Oct.  2,  1911 

3J% 

8. 

$486.50 

Jan.  8,  1911 

4  months 

Feb.  16,  1911 

5   % 

9. 

$3674.15 

Apr.  20,  1912 

60  days 

May  5,  1912 

8   % 

10. 

$398.48 

June  12,  1911 

6  months 

Aug.  8,  1911 

5J% 

11. 

$725.30 

Aug.  18,  1911 

90  days 

Sept.  3,  1911 

6   % 

12. 

$486.28 

July  31,  1911 

4  months 

Aug.  24,  1911 

9   % 

13. 

Find  the  proceeds  of  the  following 

note,  if  discounted 

July  6, 

1912,  at 

8%: 

192  PERCENTAGE 

$2576.°°  BOSTON,  MASS.,  May  26,  1912. 

Three  months  after  date,  value  received,  I  promise  to  pay 
to  Fairman  A.  Sadler,  or  order,  Two  Thousand  Five  Hundred 

Seventy-six  Dollars. 

R.  MORTIMER  BROWNING. 

14.  Find  the  proceeds  of  the  following  note,  if  discounted 
Nov.  17,  1911,  at  5%: 

$845.75  CHICAGO,  ILL.,  Oct.  18,  1911. 

Ninety  days  after  date,  value  received,   I  promise  to  pay 
Philip  A.  Smith,  or  bearer,  Eight  Hundred  Forty-five  jW  Dol- 

lars, with  interest  at  6  %. 

JAMES  J.  DAVIES. 

15.  Find  the  proceeds  of  the  following  note,  if  discounted 
May  23,  1912,  at  4J  %,  applying  Note  3: 


BALTIMORE,  MD.,  March  3,  1912. 
Four  months  after  date,  value  received,  I  promise  to  pay  to 
Allen  S.  Will,  or  order,  Seven  Hundred  Thirteen  -ffj  Dollars, 


with  interest  at  5  %. 


GEORGE  C.  ROUND. 


16.  Find  the  proceeds  of  the  following  draft,  if  accepted  July 
17,  1912,  and  paid  Aug.  20,  1912,  allowing  discount  at  4%: 

$491TV<7  NEW  ORLEANS,  LA.,  July  13,  1912. 

At  sixty  days  sight,  pay  to  the  order  of  Henry  A.  Griesemer 
&  Co.,  Four  Hundred  Ninety-one  •££$  Dollars,  value  received, 

and  charge  to  the  account  of  ~  ,T7    ~7 

GEORGE  W.  WHITESIDE. 
To  BROWN  BROS.  &  Co., 

BALTIMORE,  MD. 

17.  Find  the  proceeds  of  the  following  draft,  if  discounted 
Sept.  18,  1911,  at  4J%: 

$874fYir  NEW  YORK,  Aug.  12,  1911. 

Ninety  days  after  date,  pay  to  the  order  of  Joseph  S.  Sinclair 
&  Co.,  Eight  Hundred  Seventy-four  -ffo  Dollars,  value  received, 

and  charge  to  the  account  of 

EDMUND  BERKELEY. 
To  W.  BANKHEAD  THORNTON, 

ALEXANDRIA,  VA. 


BANK    DISCOUNT 


193 


18.  On  Oct.  15,  1911,  merchandise  to  the  amount  of  $917.65 
was  bought  at  3  months.     What  payment  on  Nov.  8,  1911,  will 
discharge  the  indebtedness,  allowing  bank  discount  at  5  %? 

19.  What  payment  on  Feb.  13,  1912,  will  discharge  a  debt  of 
$2173.85,  due  May  24,  1912,  allowing  bank  discount  at  6%? 

20.  At  5  %,  what  are  the  proceeds  of  a  60-day  note  of  $468.75, 
dated  June  17,  1911,  and  discounted  July  10,  1911? 


72. 


SOLUTION 

$647.52 

72 

129504 

453264 


TO  FIND  THE  FACE  OF  A  NOTE  TO  PRODUCE  A  GIVEN 

PROCEEDS 

ILLUSTRATIVE    EXAMPLE 

257.  What  must  be  the  face  of  a  note  payable  90  days  after 
date,  which,  if  discounted  on  day  of  issue  at  5%,  will  produce 
$647.52  as  proceeds? 

EXPLANATION.  $647.52  is  a  percentage 
of  proceeds,  or  a  product  [256,  3].  5  %  is  the 
rate  of  discount  per  annum,  or  a  multiplier, 
but  not  the  specific  rate  or  multiplier  to  pro- 
duce $647.52  proceeds  [Note,  256]. 

As    the    given    percentage    of    proceeds 

.90    453264  2      ^647-52)  equals  the  required  face  of  the  note 

or  future  worth  minus  the  interest  thereon 
for  90  days  at  5  %  per  annum,  so  its  specific 
rate  or  multiplier  must  be  the  specific  rate 
of  the  required  face  or  future  worth  (100  %) 
minus  the  specific  rate  of  discount  for  90  da. 
at  5  %  per  360  da.  [Note,  227],  or  1  -  -f£. 

Hence,  divide  the  given  percentage  of 
proceeds,  or  product  ($647.52)  by  its  specific 
rate  or  multiplier  factor  (1  -  -ff ),  or,  clear- 
ing the  divisor  of  its  fraction,  divide  72 
times  the  given  percentage  ($647.52  X  72  = 
$46621.44)  by  72  times  its  specific  rate  (72 
times  1,  or  72)  -  (72  times  fj,  or  .90),  that 
is,  by  72  -  .90,  or  71.10,  obtaining  $655.72  as 
its  multiplicand  factor,  or  the  required  face 
or  future  worth. 

VERIFICATION.  Find  the  interest  on  the  obtained  future  worth 
($655.72)  for  the  given  term  of  discount  (90  da.),  obtaining  $8.20  —  as  the 
discount,  and  subtract  the  discount  ($8.20)  from  the  obtained  future  worth 


71.10)46621.44($655.7; 
4266 
3961 
3555 
4064 
3555 
5094 
4977 

1170 
711 

459 


194  PERCENTAGE 

($655.72).     If  the  remainder  equals  the  given  proceeds  ($647.52),  the  solution 
may  be  accepted  as  mechanically  correct. 

RULE  1.  To  find  the  future  worth  or  base  in  bank  discount, 
divide  the  given  percentage  of  proceeds  by  100%  minus  the  specific 
rate  of  discount  for  the  days  of  prepayment  [Note,  227].  Or,  to 
avoid  fractions, 

2.  Multiply  the  proceeds  by  such  a  part  of  360  as  1%  is  part  of 
the  given  rate  per  annum,  and  divide  the  resulting  product  by  the 
same  part  of  860  after  diminishing  it  by  1  %  of  the  days  of  prepay- 
ment. 

NOTE.  If  the  given  rate  per  annum  of  discount  is  not  an  aliquot  of  360, 
multiply  the  given  proceeds  by  360,  and  divide  the  resulting  product  by  360 
diminished  by  as  many  times  1  %  of  the  days  of  prepayment  as  the  given  rate 
per  annum  is  times  1  %.  Thus,  in  the  111.  Ex.  ($647.52  X  360)  divided  by 
360  -  (.90  X  5),  or  by  355.50  =  $655.72.  That  is,  instead  of  dividing  the 
specific  percentage  of  proceeds  ($647.52)  by  its  specific  rate  (1  —  --f-f ),  pre- 
ferably divide  360  times  the  specific  proceeds  by  360  times  its  specific  rate,  or 
by  (360  times  1)  -  (360  times  -||). 

EXAMPLES  FOR  PRACTISE 

Find  the  face  of  a  note  which  will  produce 

1.  $648.73  proceeds,  if  discounted  85  days  before  maturity  at  5  %. 

2.  $945.85  125 6  %. 

3.  $1374.50 62 8  %. 

4.  $500 145 4J  %. 

5.  $6258.25 97 3  %. 

6.  $4000 132 7%. 

7.  $693.40 76 3|  %. 

8.  $2500 231 4  %. 

9.  $1257.85 5  mo.  16  da 6  %. 

10.  I  owe  $695.80  payable  to-day;   and  in  settlement  I  give 
my  creditor  a  60-day  note,  the  proceeds  of  which  if  discounted  at 
bank  to-day  at  6%,   will   exactly  discharge  my  indebtedness. 
What  was  the  face  of  the  note? 

11.  What  must  be  the  face  of  a  90-day  note,  which,  if  dis- 
counted on  day  of  issue  at  5  %,  will  produce  $486.25  as  proceeds? 

12.  What  should  be  the  face  of  a  note  payable  60  days  after 
date,  which  given  to  a  creditor  is  equivalent  to  a  cash  payment  of 
$500,  the  rate  of  discount  at  the  creditor's  bank  then  being  4f  %? 


UNITED    STATES   RULE   FOR    PARTIAL   PAYMENTS     195 


PARTIAL  PAYMENTS 

258.  INTRODUCTION.  Partial  payments  is  the  process  for 
finding  the  balance  due  at  a  given  date  upon  an  interest-bearing 
debt  when  one  or  more  partial  payments  thereon  have  been  pre- 
viously made.  There  are  several  methods  in  use,  only  two  of 
which  have  more  than  a  local  importance.  The  principal  of  the 
two  is  the  method  of  finding  the  balance  due  at  the  time  of  each 
successive  payment,  adopted  by  the  Supreme  Court  of  the  United 
States,  and  hence  called 


THE  UNITED  STATES  RULE  FOR  PARTIAL  PAYMENTS 
ILLUSTRATIVE  EXAMPLE 

A  note  of  $2346.75,  dated  May  16,  1909,  drawing  interest  from 
its  date  at  6  %,  has  the  following  indorsements:  paid  Oct.  23,  1909, 
$486.30;  paid  July  18,  1910,  $76.50;  paid  Feb.  25,  1911,  $500; 
paid  June  17,  1911,  $865.  What  is  the  balance  due  Nov.  2,  1911? 

SOLUTION 

Face  of  note  or  original  principal    $2346.75 

Interest  from  date  of  note,  May  16,  1909,  to  date  of  first  pay- 
ment, Oct.  23,  1909  (time  by  30-day  method  =  157  da.),  is 
$61.41;  which  is  deducted  from  the  first  payment  ($486.30) 
leaving  a  balance  of  $424.89  to  apply  to  the  reduction  of  the 
principal 424.89 

Balance  due  after  making  the  first  payment $1921.86 

Interest  of  unpaid  principal  ($1921.86)  from  its  date,  Oct.  23, 
1909,  to  the  date  of  the  next  following  payment,  July  18,  1910 
(265  da.),  is  $84.88,  or  $8.38  more  than  the  second  payment 
$76.50).  This  unpaid  interest  ($8.38)  is  carried  forward  to  the 
accrued  interest  at  the  date  of  the  third  payment,  to  find  the 
total  interest  then  due. 

Interest  of  same  principal  ($1921.86)  from  the  date  of  the 
last  considered  payment,  July  18,  1910,  to  the  date  of  the  next 
following  payment,  Feb.  25,  1911  (217  da.),  is  $69.51.  To  this, 
add  the  unpaid  interest  from  the  last  operation  ($8.38),  making 
the  total  unpaid  interest  on  Feb.  25,  1911,  equal  $77.89,  which 
is  deducted  from  the  third  payment  ($500),  leaving  a  credit 
balance  of  $422.11  to  apply  to  the  reduction  of  the  unpaid 
principal 422.11 

Balance  due  after  making  the  third  payment $1499.75 


196  PERCENTAGE 

Balance  due  after  making  the  third  payment $1499.75 

Interest  of  new  principal  ($1499.75)  from  its  date,  Feb.  25, 
1911,  to  the  date  of  the  next  following  payment,  June  17,  1911 
(112  da.),  is  $28,  which  is  deducted  from  the  fourth  payment 
($865),  leaving  a  credit  balance  of  $837  to  apply  to  the  reduc- 
tion of  the  unpaid  principal  837. 

Balance  due  after  making  the  fourth  payment    $   662.75 

Interest  of  new  principal  ($662.75)  from  its  date,  June  17, 
1911,  to  the  date  of  final  settlement,  Nov.  2,  1911  (135  da.),  is 14.91 

Balance  due  at  date  of  final  settlement,  Nov.  2,  1911    $  677.66 

RULE.  1.  From  the  first  payment,  deduct  the  accrued  interest 
upon  the  principal  to  the  date  of  said  payment,  and  diminish  the 
principal  by  the  remainder,  if  any,  to  find  the  unpaid  principal. 

2.  Deduct  from  the  second  payment  the  accrued  interest  upon  the 
unpaid  principal  to  the  date  of  the  second  payment,  and  diminish 
the  unpaid  principal  by  the  remainder,  if  any,  to  find  how  much  of 
the  interest-bearing  principal  is  still  unpaid. 

3.  //  the  interest  due  at  the  date  of  any  payment  should  prove  to 
be  greater  than  said  payment,  deduct  the  payment  from  the  interest 
and  add  the  unpaid  interest  to  the  accrued  interest  for  the  next  fol- 
lowing interval  to  find  the  total  unpaid  interest  at  the  end  of  the  next 
following  interval;  then  proceed  in  the  usual  manner. 

4.  Continue  this  process  consecutively  until  every  payment  has 
been  considered. 

5.  The  final  balance  will  be  the  balance  due  at  the  last  payment, 
increased  by  the  interest  thereon  from  the  date  of  the  last  payment  to 
the  date  of  the  final  settlement. 

NOTE  1.  By  the  U.  S.  rule,  any  payment  is  first  applied  to  the  dis- 
charge of  the  accrued  interest  to  the  date  of  such  payment,  and  the  remainder 
of  the  payment,  if  any,  is  then  applied  to  the  reduction  of  the  interest-bearing 
principal;  and  any  deficiency  of  a  payment  to  discharge  the  accrued  interest 
is  carried  forward  to  the  interest  for  the  next  following  interval  to  find  the 
total  unpaid  interest  at  the  end  of  the  next  following  interval. 

NOTE  2.  The  unpaid  balance  of  the  principal  at  any  intermediate  settle- 
ment should  never  exceed  the  immediately  preceding  unpaid  balance.  Only 
unpaid  interest  could  have  caused  it  to  be  greater;  and  to  compute  interest 
upon  such  a  balance  would  be  virtually  to  compute  interest  upon  the  unpaid 
interest  it  contains  (compound  interest)  to  which  the  U.  S.  Rule  is  opposed. 

NOTE  3.  The  Supreme  Court  uniformly  employs  the  30-day  method 
[163;  Note  4,  233]  in  finding  the  time  between  two  dates. 


MERCHANTS'    RULE    FOR   PARTIAL    PAYMENTS        197 

EXAMPLES  FOR  PRACTISE 

1.  A  note  of  $3750,  dated  Sept.  26,  1908,  drawing  interest 
from  date  at  6%,  has  the  following  payments  indorsed  thereon: 
May  9,  1910,  $575;  Jan.  17,  1911,  $750.     Find  the  balance  due 
July  21,  1911. 

2.  A  note  of  $2975,  dated  Aug.  20,  1909,  drawing  interest  at 
5%,  was  credited  with  the  following  payments:    Nov.  12,  1910, 
$473.85;  Mch.  21,  1911,  $678.35;  Sept.  9,  1911,  $296.80.     What 
was  the  balance  due  Dec.  23,  1911? 

3.  What  was  the  balance  due  Apr.  12,  1912,  on  a  note  of 
$816.25,   dated  Oct.   18,    1908,   and  drawing  interest  at  4J  %? 
Payments:  June  24,  1909,  $256.25;  Mch.  8,  1910,  $176.80;  Nov. 
20,  1910,  $156.25;  May  25,  1911,  $150. 

4.  A  note  of  $1285.60,  dated  Oct.  25,  1907,  drawing  interest  at 
6%,  has  the  following  indorsements:   Mch.  13,  1909,  $200;   Feb. 
12,  1910,  $352.75;    Dec.  29,  1910,  $148;    Apr.  20,  1911,  $275, 
What  was  the  balance  due  July  8,  1911? 

6.  A  note  of  $6500,  dated  May  23, 1910,  drawing  interest  at  5  %, 
was  indorsed  as  follows:  Nov.  17,  1910,  $125;  June  20, 1911,  $400; 
Feb.  16,  1912,  $100.  What  was  the  balance  due  Sept.  12,  1912? 

6.  A  note  of  $5000,  dated  Feb.  17,  1910,  drawing  6  %  interest, 
was  paid  in  full  on  Oct.  10,  1912.     What  was  the  final  payment,  if 
the  following  partial  payments  had  been  previously  made :  Dec.  19, 
1910,  $200;  Sept.  16,  1911,  $200;  May  25,  1912,  $200? 

7.  I  have  a  note  of  $658.25,  drawing  interest  at  6  %,  and  dated 
July  24,  1910,  upon  which  I  have  received  the  following  payments: 
Oct.  12,  1910,  $45;   Apr.  28,  1911,  $10.     What  was  the  balance 
due  Sept.  16,  1911? 

8.  A  note  of  $956.25,  dated  Jan.  16,  1912,  drawing  6  %  interest, 
had  a  payment  of  $375.80,  dated  May  14,  1912,  indorsed  thereon. 
On  Aug.  12,  1912,  the  maker  of  the  note  made  a  second  payment  of 
$250,  and  gave  a  new  interest-bearing  note  for  the  remainder. 
What  was  the  face  of  the  new  note? 

MERCHANTS'  RULE  FOR  PARTIAL  PAYMENTS 

259.  The  merchants'  rule  is  so  called  because  it  conforms  to 
the  usage  of  merchants  in  obtaining  the  balances  of  all  accounts 
which  contain  past-due  (that  is,  interest-bearing)  items;  as  hi 


198  PERCENTAGE 

finding  the  "cash  balance"  of  any  account  current  [299],  or  as 
in  fmo5ng  the  balance  due  at  a  given  date  "per  average"  [296]. 

ILLUSTRATIVE  EXAMPLE 

A  debt  of  $478.30,  drawing  interest  at  6  %  from  May  14,  1911, 
was  credited  with  the  following  partial  payments:  July  7,  1911, 
$178.35;  Sept.  20,  1911,  $200;  Oct.  25,  1911,  $75.50.  Find  the 
balance  due  Dec.  3,  1911. 

SOLUTION 

The  principal,  or  debt     $478.30 

Interest  of  $478.30  from  May  14  to  Dec.  3  (exact  time,  203  da.) 16.18 

Indebtedness  at  settlement  if  no  payments  had  been  made    $494.48 

First  payment  on  July  7,  1911    $178.35 

Interest  of  $178.35  from  July  7,  to  Dec.  3  (149  da.)   4.43 

Second  payment  on  Sept.  20,  1911     200.00 

Interest  of  $200  from  Sept.  20  to  Dec.  3  (74  da.)    2.47 

Third  payment  on  Oct.  25,  1911     75.50 

Interest  of  $75.50  from  Oct.  25  to  Dec.  3  (39  da.) .49          461.24 

Debit  balance  on  Dec.  3     $  33.24 

RULE.  1.  Increase  the  debt  by  its  accrued  interest  from  the  date 
it  commenced  to  draw  interest  to  the  date  of  final  settlement. 

2.  Find  the  interest  of  each  payment  from  the  date  of  its  payment 
to  the  date  of  final  settlement. 

3.  Subtract  the  sum  of  the  payments  and  their  accrued  interest 
from  the  sum  of  the  debt  and  its  accrued  interest. 

NOTE  1.  The  above  is  simply  the  common  practise  among  merchants  of 
subtracting  the  total  credits  at  any  given  date  from  the  total  debits  at  that 
same  date,  to  find  the  debit  balance  at  that  date. 

EXAMPLES  FOR  PRACTISE 

Direction:  In  the  following  problems,  find  the  exact  time  in  days  [Note 
5,  233],  though  usage  in  this  respect  is  not  uniform  in  all  cities  of  the  United 
States. 

1.  What  was  the  balance  due  Sept.  28,  1912,  on  a  note  of 
$792.50,  dated  May  20,  1912,  drawing  interest  from  date  at  6  %, 
and  containing  the  following  indorsements:   July  12,  1912,  $150; 
Aug.  8,  1912,  $200. 

2.  A  note  of  $1873.25,  dated  Nov.  17,  1910,  drawing  interest 
from  date  at  5%,  was  credited  with  the  following  payments: 


MERCHANTS'    RULE   FOR   PARTIAL    PAYMENTS        199 

Jan.  28,  1911,  $375.80;   May  16,  1911,  $468.25;   June  25,  1911, 
$268.10.     What  was  the  balance  due  Aug.  12,  1911? 

3.  Find  balance  due  Dec.  28,  1912,  on  a  bill  of  merchan- 
dise amounting  to  $398.75,  purchased  Feb.  16,  1912,  on    two 
months'  credit,  computing  interest  at  6%  from  the  due  date  of 
the  purchase.     Payments:  May  13,  1912,  $85.30;  July  27,  1912, 
$49.75;  Oct.  3,  1912,  $158.60. 

4.  At  6  %,  find  balance  due  Jan.  6,  1912,  on  a  bill  of  groceries 
amounting  to  $678.35,  purchased  Mch.  3,  1911,  on  three  months' 
credit  without  interest.     Payments:  Apr.  20, 1911,  $275;  July  26, 
1911,  $150;  Nov.  16,  1911,  $145.80. 

SUGGESTION.  If  payments  are  made  before  a  non-interest  bearing  pur- 
chase falls  due  (as  in  the  payment  of  $275  on  April  20,  1911),  first  find  the 
balance  due  at  maturity  by  subtracting  the  prepayments  plus  their  accrued 
interest  to  the  due  date  (that  is,  plus  their  true  discount),  from  the 
non-interest  bearing  purchase,  to  find  the  debit  balance  at  maturity,  which 
draws  interest  from  the  due  date  of  the  purchase;  after  which  proceed  as  usual 
with  the  postpayments,  if  any. 

6.  At  6  %,  find  balance  due  Nov.  25,  1910,  upon  a  purchase  of 
$846.25  on  May  18,  1910,  at  90  days'  credit  without  interest. 
Payments:  June  10,  1910,  $175.30;  July  14,  1910,  $200;  Sept.  28, 
1910,  $250. 

6.  At  8  %,  find  balance  due  Jan.  29,  1912,  upon  a  purchase  of 
$328.65  on  Apr.  7,  191 1 ,  at  4  months  without  interest.     Payments : 
May  18,  1911,  $75.40;  June  23,  1911,  $183.25;  July  10,  1911, 
$45.80. 

7.  At  9  %,  find  balance  due  Feb.  20,  1912,  upon  a  purchase  of 
$419.25  on  July  16,  1911,  at  30  days  without  interest.     Payments: 
Aug.  15,   1911,  $175.90;    Oct.   10,   1911,  $85.30;    Jan.  6,  1912, 
$65.70. 

8.  A  purchase  of  $956.30  on  April  13,  1910,  and  payable  3 
months  after  date,  had  the  following  payments  credited  thereon: 
May  20,  1910,  $200;   Sept.  13,  1910,  $300;   Oct.  25,  1910,  $150. 
Find  balance  due  Dec.  13,  1910,  at  6  %. 

9.  At  6%,  find  balance  due  June  12,  1913,  upon  a  purchase 
of  $715.25  on  Nov.  20,  1912,  at  60  days  without  interest.     Pay- 
ments;  Dec.  10,  1912,  $158.60;    Jan.  19,  1913,  $175;    Mch.  3, 
1913,  $148.65;  Apr.  25,  1913,  $125.90;  May  1,  1913,  $100. 


200  PERCENTAGE 


COMMISSION  AND  BROKERAGE 

260.  INTRODUCTION.   Commission  or  brokerage  is  the  appli- 
cation of  the  principles  of  percentage,  to  any  business  transac- 
tion performed  by  one  person,  called  an  agent,  for  the  benefit  of 
another  person,  called  a  principal.     Thus,  it  may  be  inconvenient 
as  well  as  unnecessarily  expensive  to  travel  a  long  distance  to  the 
best  market  for  buying  a  particular  kind  of  merchandise,  or  one 
may  not  possess  the  requisite  skill  or  knowledge  of  local  condi- 
tions to  buy  or  sell  under  the  most  favorable  circumstances;   or 
one  may  not  have  access  to  the  floor  of  the  "  Exchange"  or  " Cham- 
ber of  Commerce,"  or  " Board  of  Trade"  at  which  the  fluctuations 
in  prices  are  posted;   and  it  may  therefore  be  so  highly  desirable 
to  secure  the  services  of  one  who  possesses  these  advantages,  as 
to  justify  the  payment  of  a  commission,  or  remuneration,  to  him 
for  the  service  rendered. 

NOTE.  What  has  been  said  applies  not  only  to  merchandise,  but  to  real 
estate,  or  stocks,  or  bonds,  which  are  bought  or  sold  by  an  agent;  or  to 
collections  of  money;  or  to  any  other  authorized  special  mercantile  service 
rendered  by  an  agent  to  his  principal. 

261.  Commission  or  brokerage  is  a  charge  made  by  an  agent 
for  buying  or  selling  merchandise,  or  real  estate,  or  other  property, 
or  for  collecting  or  disbursing  moneys. 

NOTE  1.  The  amount  of  commission  will  vary  with  the  extent  of  the  ser- 
vice rendered,  and  in  buying  or  selling  is  usually,  estimated  at  a  certain  % 
of  the  sum  paid  by  the  agent  for  the  merchandise  bought  (its  prime  cost), 
or  received  by  the  agent  for  the  merchandise  sold  (its  gross  proceeds).  In 
many  instances,  however,  it  is  computed  at  a  certain  price  per  unit,  irrespec- 
tive of  the  sum  paid  or  received  for  it,  as  grain  per  bushel,  flour  per  barrel, 
cotton  per  bale,  leaf  tobacco  per  hogshead,  stocks  per  share,  small  fruits  per 
crate,  etc. 

NOTE  2.  If  a  consignment  is  sold  on  credit,  in  addition  to  the  com- 
mission for  selling,  a  certain  %  of  the  credit  sale,  called  guaranty,  is  charged 
by  the  agent  for  assuming  responsibility  for  its  final  payment. 

NOTE  3.  Merchandise  forwarded  to  an  agent  to  be  sold  is  called  a  con- 
signment, the  person  who  sends  the  merchandise  being  called  the  consignor 
or  shipper,  and  the  agent  to  whom  it  is  sent,  the  consignee. 

262.  The  gross  proceeds  of  a  sale  is  the  total  sum  received  by 
the  agent  from  the  buyer  of  the  consignment  before  deducting  the 
expenses  incidental  to  the  sale. 


COMMISSION    AND    BROKERAGE  201 

NOTE.  After  all  the  expenses  of  the  sale  have  been  deducted  from  the 
gross  proceeds,  the  remainder  is  called  the  net  proceeds  of  the  sale. 

263.  The  prime  cost  of  a  purchase  is  the  sum  paid  by  the 
agent  to  the  seller  thereof,  and  does  not  include  any  expense  inci- 
dental to  the  purchase. 

NOTE.  After  the  incidental  expenses  have  been  added  to  the  original 
purchase  money,  the  result  is  called  the  gross  cost. 

264.-  Identification  of  terms.  1.  The  prime  cost  of  a  purchase, 
or  the  gross  proceeds  of  a  sale,  is  the  base  or  multiplicand. 

2.  The  specific  rate  of  any  given  or  required  percentage  of  com- 
mission, of  brokerage,  of  guaranty,  of  net  proceeds  of  a  sale,*  or  of 
gross  cost  of  a  purchase,  f  is  the  rate  or  multiplier. 

3.  The  commission,  the  brokerage,  the  guaranty,  the  net  pro- 
ceeds of  a  sale,  or  the  gross  cost  of  a  purchase,  is  a  percentage  or 
product. 

*  NOTE  1.  As  the  net  proceeds  of  a  sale  equal  the  gross  proceeds  or  base 
minus  all  the  expenses,  so  the  specific  rate  of  net  proceeds  must  equal  the  rate 
of  gross  proceeds  or  base  (100  %)  minus  the  rate  of  all  the  expenses. 

t  NOTE  2.  As  the  gross  cost  of  a  purchase  equals  the  prime  cost  or  base 
plus  all  the  expenses,  so  the  specific  rate  of  gross  cost  must  equal  the  rate  of 
prime  cost  or  base  (100  %)  plus  the  rate  of  all  the  expenses. 

NOTE  3.  If  an  expense  incidental  to  a  purchase  or  to  a  sale  is  a  non- 
percentage,  that  is,  if  it  has  not  been  computed  at  a  certain  %  of  the  base,  as 
drayage,  freight,  storage,  cooperage,  advertising,  etc.,  any  net  proceeds  of  a 
sale  or  gross  cost  of  a  purchase  which  includes  one  or  more  such  non-percentage 
or  non-product  charges,  must  itself  be  a  non-percentage  or  non-product,  and 
can  therefore  have  no  specific  rate  or  multiplier.  Hence,  add  all  the  non-per- 
centage charges  to  a  given  non-percentage  net  proceeds,  or  subtract  them  from 
a  given  non-percentage  gross  cost,  to  reduce  it  to  a  percentage  or  product, 
after  which  apply  the  principles  of  percentage  in  the  usual  manner. 

EXAMPLES   FOR  PRACTISE 

1.   If  the  prime  cost  of  a  purchase  is  $625.80  and  the  rate  of 
commission  for  buying  is  3  %,  what  is  the  commission? 

Prime  cost  or  base  X  specific  rate  of  commission  =  commission. 

Find  the  commission,  if  the 

x   2.   Prime  cost  is  $378.25,  and  the  rate  of  commission  3  %. 
^  3.   Gross  proceeds  are  $234.70,  and  the  rate  of  commission  2  %. 
4.   Prime  cost  is  $1348.62,  and  the  rate  of  commission  2|  %. 


202  PERCENTAGE 

6.   Gross  proceeds  are  $6358.20,  and  the  rate  of  brokerage  f  %. 

6.  Prime  cost  is  $5786.90,  and  the  rate  of  brokerage  f  %. 

7.  If  the  gross  proceeds  of  a  sale  are  $875.30,  the  rate  of  com- 
mission 2  %,  and  the  drayage  $3.75,  what  are  the  net  proceeds? 

Base  —  (base  X  rate   of   com.)  —  other   charges  =  net  proceeds. 

Find  the  net  proceeds  of  a  sale,  if  the  gross  proceeds  are 

8.  $586.75,  the  commission  3  %,  and  the  other  expenses  $6.35. 

9.  $2783.60,  the  com.  1J%,  and  the  other  expenses  $37.20. 

10.  $8625,  the  brokerage  T3ir  %,  and  the  other  expenses  $136.50. 

11.  $7594.16,  the  brok.  f  %,  and  the  other  expenses  $213.75. 

12.  If  the  prime  cost  of  a  purchase  is  $763.42,  the  rate  of  com- 
mission li  %,  and  the  other  charges  $35.90,  what  is  the  gross  cost? 

Base  +  (base  X  rate  of  com.)  +  other  charges  =  gross  cost. 

Find  the  gross  cost  of  a  purchase,  if  the  prime  cost  is 

13.  $853.75,  the  com.   If  %,  and  the  other  charges  $25.60. 

14.  $2346.50,  the  com.  li%,  and  the  other  charges  $17.25. 

15.  $18250.75,  the  brok.  |  %,  and  the  other  charges  $362.75. 

16.  $9580,  the  brokerage  f%,  and  the  other  charges  $237.15. 

17.  If  the  prime  cost  of  a  purchase  is  $687.50,  and  the  gross 
cost  $701.25,  what  is  the  rate  %  of  commission? 

EXPLANATION.  Divide  the  specific  percentage  or  product  of  commission 
($701.25  —  $687.50  =  $13.75,  commission)  by  its  base  or  multiplicand  factor 
($687.50),  obtaining  2  %  as  the  specific  rate  of  com.,  or  multiplier  factor. 

Find  the  rate  %  of  commission,  if  the 

18.  Gross  proceeds  are  $685.75,  and  the  commission  $27.43. 

19.  Prime  cost  is  $468.50  and  the  commission  $9.37. 

20.  Gross  pro.  are  $768.35,  and  the  commission  $23.05  [205,  a]. 

21.  Prime  cost  is  $18750,  and  the  commission  $70.31  [205,  c]. 

22.  Gross  proceeds  are  $672.85,  and  the  net  proceeds  $645.94. 

23.  Prime  cost  is  $296.45,  gross  cost  $312.52,  and  drayage  $1 .25. 

24.  What  is  the  prime  cost  of  a  purchase,  if  the  commission  is 
$20.57,  and  the  rate  of  commission  3  %? 

Given  percentage  -f-  its  specific  rate  =  its  specific  base. 

Find  the  gross  proceeds  or  the  prime  cost,  if  the 

25.  Commission  for  buying  is  $10.69,  and  the  rate  of  com.  2 \  %. 

26.  Commission  for  selling  is  $27.48,  and  the  rate  of  com.  3  %. 


COMMISSION   AND    BROKERAGE  203 

27.  Brokerage  for  buying  is  $42.50,  and  the  rate  of  brok.  J  %. 

28.  Brokerage  for  selling  is  $37.60,  and  the  rate  of  brok.  f  %. 

29.  Net  proceeds  are  $947.36,  and  the  rate  of  commission  3  %. 

30.  Net  proceeds  are  $658.56,  and  the  rate  of  commission  2  %. 

31.  Gross  cost  is  $5018.75,  and  the  rate  of  brokerage  f  %. 

32.  Gross  cost  is  $972.36,  and  the  rate  of  commission  1 J  %. 

33.  Net  proceeds  are  $624.25,  the  com.  3  %,  the  drayage  $6.25, 
(Net  proceeds  +  non-percentage  charges)  -5-  specific  rate  =  gross  proceeds. 

34.  Net  pro.  are  $487.25,  com.  2J%,  other  expenses  $1.50. 

35.  Net  pro.  are  $855.24,  com.  2  %,  other  charges  $2.75. 

36.  Net  pro.  are  $446.85,  com.  4%,  other  charges  $3.15. 

37.  Gross  cost  is  $751.09,  com.  3%,  other  charges  $1.25. 
(Gross  cost  —  non-percentage  charges)  -5-  specific  rate  =  prime  cost. 

38.  Gross  cost  is  $269.52,  commission  2  %,  other  expenses  75  f. 

39.  Gross  cost  is  $585.45,  commission  1 }  %,  freight  $2.25. 

40.  Gross  cost  is  $874.41,  commission  If  %,  drayage  $7.50. 

41.  If  the  net  proceeds  are  $376.35,  and  the  rate  of  commis- 
sion 2J  %,  what  is  the  commission? 

42.  If  the  gross  cost  is  $425.39  and  the  rate  of  commission 
3J  %,  what  is  the  commission? 

43.  An  agent  received  $375.18  to  invest  in  wheat.     If  his  rate 
of  commission  was  3  %,  what  was  his  commission? 

44.  An  agent  remitted  $462.95  to  his  principal  as  the  net  pro- 
ceeds of  a  sale.     If  the  agent's  rate  of  commission  was  1 J  %,  what 
were  the  gross  proceeds  of  the  sale? 

45.  An  agent  sold  9873  pounds  of  sugar  at  4f  cents  per  pound, 
charged  If  %  commission  for  selling,  and  $2.90  for  other  expenses. 
What  were  the  net  proceeds  of  the  sale? 

46.  An  agent  bought  5278  pounds  bacon  at  13|  cents  per 
pound,  charged  2J  %  commission  for  buying,  and  75  cents  for 
drayage.     What  sum  should  his  principal  remit  in  settlement? 

47.  An  agent  received  $348.25  to  invest  in  potatoes.     If  his 
commission  was  3%,  how  many  barrels  could  he  buy  at  $2.15 
per  barrel,  and  what  was  the  unexpended  balance,  if  any? 

SUGGESTION.     In  business  practise,  no  fraction  of  a  barrel  is  bought; 
nor  is  the  agent  entitled  to  a  commission  on  any  possible  unexpended  balance 


204  PERCENTAGE 

of  a  remittance.  Hence,  divide  the  remittance  ($348.25)  by  the  gross  cost 
of  1  barrel,  to  find  the  number  of  barrels  which  can  be  bought  with  the  remit- 
tance, and  regard  any  remainder  from  the  division  as  an  unexpended  balance. 

48.  An  agent  sold  495  bushels  wheat  at  $1.03  per  bushel,  and 
with  the  proceeds  bought  flour  at  $5.95  per  barrel,  the  commission 
for  buying  and  selling  being  2  %.     How  many  barrels  of  flour  did 
the  agent  buy,  and  what  was  the  unexpended  balance? 

49.  An  agent  received  $298.40  to  invest  in  wheat.     If  his  com- 
mission was  2J  %,  how  many  bushels  could  he  buy  at  98  cents  per 
bushel? 

SUGGESTION.  Unlike  merchandise  bought  by  the  barrel,  in  buying  grain 
there  is  no  unexpended  balance;  but  the  remainder  from  the  division  should 
be  expressed  as  a  fraction,  the  denominator  of  which  equals  the  number  of 
pounds  in  one  bushel  of  the  particular  kind  of  grain  under  consideration 
[Note  1,  142].  In  this  problem,  to  obtain  pounds  or  60ths  of  a  bushel,  mul- 
tiply the  remainder  from  the  division  by  60,  and  divide  the  resulting  product 
by  the  original  divisor  [86]. 

50.  An  agent  sold  75  barrels  of  flour  at  $6.25  per  barrel  net 
cash,  and  50  barrels  at  $6.40  per  barrel,  payable  in  3  months. 
His  charges  were  2  %  for  selling,  \\%  for  guaranty,  and  $2.50  for 
other  expenses.     What  sum  should  he  remit  to  his  principal  in 
settlement? 

51.  At  what  price  shall  an  agent  be  ordered  to  buy  flour  at 
2  %  commission  that,  after  allowing  5  cents  per  barrel  for  freight 
and  2  cents  per  barrel  for  dray  age,  the  flour  can  be  sold  at  $5.94 
per  barrel  and  net  20  %  profit? 

52.  A  broker  sold  5600  bushels  of  wheat  at  $1.09  per  bushel, 
and  charged  \i  per  bushel  for  selling.     What  were  the  net  pro- 
ceeds of  the  sale? 

53.  An  agent  sold  28356  pounds  leaf  tobacco  at  $15.60  per 
hundredweight,  charged  1 J  %  commission  for  selling,  and  $7.50 
for  other  expenses.     What  were  the  net  proceeds  of  the  sale? 

54.  An  agent  bought  18725  pounds  wheat  at  $1.15  per  bushel, 
charged  2%  for  commission,  25  cents  per  load  of  36  bushels  for 
drayage,  and  6  cents  per  cwt.  for  freight  prepaid.     What  sum 
should  his  principal  remit  to  him  in  settlement? 

55.  An   agent   received   $348.72   to   invest   in   onions.     His 
charges  were  3  %  for  commission,  2  cents  per  barrel  for  drayage, 


STOCKS   AND    BONDS  205 

5  cents  per  barrel  for  freight,  and  $7.60  for  other  expenses.  How 
many  barrels  of  onions  could  he  buy  at  $1.95  per  barrel,  and  what 
was  the  unexpended  balance? 

SUGGESTIOX.  First  diminish  the  remittance  by  the  sum  of  the  charges 
which  were  not  estimated  per  barrel,  and  divide  the  remainder  by  the  prime 
cost  per  barrel  increased  by  the  charges  which  were  directly  or  indirectly 
estimated  per  barrel.  The  quotient  will  denote  the  number  of  barrels.  Re- 
gard any  remainder  from  the  division  as  an  unexpended  balance. 

66.  An  agent  received  $427.95  to  invest  in  apples.  He  bought 
the  apples  at  $2.75  per  barrel  and  charged  3  %  commission  for 
buying,  25  cents  per  load  of  10  barrels  for  drayage,  6  cents  per 
barrel  for  freight,  and  $2.95  for  advertising.  How  many  barrels 
of  apples  could  the  agent  buy  with  the  remittance,  and  what  was 
the  unexpended  balance? 

SUGGESTION.  Divide  the  drayage  per  load  by  the  number  of  barrels  per 
load  to  find  the  drayage  per  barrel;  and  proceed  as  in  Ex.  55  to  find  the  num- 
ber of  barrels.  Then  multiply  the  drayage  per  barrel  for  full  loads  by  the 
number  of  barrels  which  the  last  load  lacks  of  being  a  full  load,  to  find  the 
extra  drayage  (in  excess  of  the  former  estimated  drayage  per  barrel  for  full 
loads)  incident  to  the  last  incomplete  load;  and  deduct  it  from  what  would 
otherwise  have  been  the  unexpended  balance,  to  find  the  correct  unexpended 
balance. 

57.  An  agent  received  $485.30  to  invest  in  corn.     His  charges 
were  2  %  for  commission,  40  cents  per  load  of  50  bushels  for  dray- 
age, and  5  cents  per  cwt.  for  freight  prepaid.     How  many  bushels 
and  pounds  (56ths  of  a  bushel)  did  he  buy  at  65  cents  per  bushel? 

58.  An  agent  received  $472.10  to  invest  hi  potatoes.     He 
charged  3%  commission  for  selling,  35  cents  per  load  of  12  barrels 
for  drayage,  4  cents  per  barrel  for  freight,  and  $2.40  for  other 
expenses.      How  many  barrels  of  potatoes  could  the  agent  buy 
at  $1.95  per  barrel,  and  what  was  the  unexpended  balance? 

STOCKS  AND  BONDS 

265.  INTRODUCTION,  (a)  A  corporation  is  an  association  of 
persons  which  is  authorized  by  a  special  legal  instrument  called 
a  charter  to  transact  business  under  certain  limitations  and  obli- 
gations as  if  it  were  a  person. 

(6)  The  stock  of  a  corporation  is  so  much  of  its  capital  as  is 
represented  by  its  shares.  A  share  of  a  corporation  is  one  of  the 


206  PERCENTAGE 

equal  parts  into  which  its  capital  stock  is  divided.  As  a  con- 
venience in  expressing  its  market  value,  in  declaring  dividends 
or  assessments,  in  computing  the  premium  or  discount  at  which  it 
is  bought  or  sold,  the  capital  stock  is  usually  divided  into  as  many 
shares  of  $100  each,  or  of  some  multiple  or  aliquot  thereof,  as  will 
equal  the  required  capital.  The  owner  of  one  or  more  of  these 
shares  is  called  a  stockholder,  and  the  instrument  issued  by  a  cor- 
poration to  each  stockholder  to  certify  how  many  shares  he  holds 
is  called  a  stock  certificate. 

(c)  The  bonds  of  a  corporation  are  so  much  of  its  capital  as 
is  represented  by  its  bonded  indebtedness.     For  convenience  in 
making  interest  calculations,  etc.,  each  bonded  loan  is  usually 
divided  into  as  many  equal  parts  of  $100  each,  or  of  some  mul- 
tiple or  aliquot  thereof,  as  will  equal  the  amount  of  the  loan. 
Those  who  take  one  or  more  of  these  equal  parts  of  a  loan  are 
called  bondholders. 

(d)  Computations  in  stocks  or  bonds  are  made  in  accordance 
with  the  general  principles  of  percentage,  the  par  or  face  value  of 
the  stock  or  bond  being  very  naturally  regarded  as  the  basis  of 
such  calculations  [207].     Thus,  if  a  corporation  prospers  and  in 
consequence  is  paying  better  returns  upon  its  par  or  face  value 
than  the  same  sum  of  money  invested  in  other  equally  safe  ways, 
its  stock  will  sell  at  such  a  %  of  premium  (increase)  on  its  par 
value  as  will  approximately  equal  this  difference ;  or  if  the  returns 
on  its  par  or  face  value  are  not  so  good  as  those  of  the  same  sum 
of  money  invested  in  other  equally  safe  ways,  its  stock  is  not  so 
desirable,  and  will  only  sell  at  such  a  %  of  discount  (decrease) 
on  its  par  value  as  will  approximate  the  deficiency;   etc.;   etc. 

266.  The  par  value  is  the  value  specified  upon  the  face  of 
stocks  or  of  bonds. 

267.  The  market  value  is  the  value  of  stocks  or  of  bonds  upon 
the  stock  exchange  or  stock  market. 

NOTE.  If  the  market  value  exceeds  the  par  value,  stocks  or  bonds  are 
said  to  be  above  par  or  at  a  premium;  and  if  the  market  value  is  less  than  the 
par  value,  stocks  or  bonds  are  said  to  be  below  par  or  at  a  discount.  The 
amount  of  premium  or  discount  per  share  is  the  difference  between  its  par 
value  and  its  market  value. 

268.  A  dividend  is  that  part  of  the  earnings  of  a  corporation 
which  is  divided  among  its  stockholders. 

NOTE.  Occasionally,  though  rarely,  special  dividends  have  been  declared 
by  corporations  from  other  sources  than  their  earnings,  as  from  loans  nego- 
tiated for  that  purpose,  or  from  the  sale  of  an  extra  issue  of  stock,  etc. 


STOCKS   AND   BONDS  207 

269.  An  assessment  is  a  sum  levied  upon  the  stockholders  of 
a  corporation  to  meet  its  losses,  or  its  extraordinary  expenditures 
for  betterments,  etc. 

270.  A  quotation  is  the  market  value  of  $100  par  value  of 
stocks  or  of  bonds. 

NOTE.  Mining  stocks  are  mostly  quoted  at  so  much  per  share,  without 
reference  to  their  par  value.  In  some  stock  exchanges,  all  stocks,  the  par  value 
per  share  of  which  is  less  than  $100,  are  thus  quoted.  Bonds  of  all  denomina- 
tions are  quoted  at  the  market  value  of  each  $100  of  their  face  value. 

271.  A  stock-broker  is  a  person  whose  business  it  is  to  buy  or 
sell  stocks  pr  bonds  for  others. 

NOTE.  The  charge  of  a  broker  for  buying  or  selling  stocks  or  bonds,  is 
called  brokerage.  Brokerage  is  estimated  at  a  certain  %  of  the  par  value,  or 
at  so  much  per  share. 

272.  Identification  of  terms.     1.    The  par  value  of  stocks  or  of 
bonds  is  the  base  or  multiplicand  understood  when  no  other  base  is 
specifically  mentioned. 

2.  The  specific  rate  of  premium,  of  discount,  of  dividend,  of 
assessment,  of  market  value,  of  prime  or  gross  cost  when  buying,  or 
of  net  or  gross  proceeds  when  selling,  is  the  rate  or  multiplier. 

3.  The  premium,  the  discount,  the  dividend,  the  assessment,  the 
market  value,  the  prime  or  gross  cost  of  a  purchase,  the  net  or  gross 
proceeds  of  a  sale,  is  a  percentage  or  product. 

NOTE  1.  A  quotation  is  the  specific  rate  %  of  market  value.  Hence,  as 
stocks  and  bonds  are  bought  and  sold  at  their  market  value,  the  quotation 
should  be  regarded  as  the  specific  rate  %  of  prime  cost  when  buying,  or  the 
specific  rate  %  of  gross  proceeds  when  selling.  The  usual  symbol  of  a  rate  (  %) 
being  omitted  from  the  quotation,  the  learner  is  cautioned  against  the  common 
error  of  confounding  a  quotation  which  is  a  rate  or  multiplier  in  its  relationship 
to  other  terms,  with  a  percentage  which  is  a  product  in  its  relationship. 

NOTE  2.  If  there  is  no  brokerage,  the  quotation  or  specific  rate  of  prime 
cost  when  buying  (Note  1)  will  also  be  the  specific  rate  of  gross  cost  or  of 
investment;  but  if  brokerage  is  included  in  the  transaction,  the  rate  of  broker- 
age must  be  added  to  the  quotation  or  specific  rate  of  prime  cost  to  obtain 
the  specific  rate  of  gross  cost.  Reversely,  the  rate  of  brokerage  must  be  sub- 
tracted from  any  obtained  rate  of  gross  cost  to  find  a  required  quotation  or 
rate  of  prime  cost. 

NOTE  3.  If  there  is  no  brokerage,  the  quotation  or  specific  rate  of  gross 
proceeds  when  selling  (Note  1)  will  also  be  the  specific  rate  of  net  proceeds; 


208  PERCENTAGE 

but  if  brokerage  is  included  in  the  transaction,  the  rate  of  brokerage  should  be 
deducted  from  the  quotation  or  specific  rate  of  gross  proceeds  to  find  the 
specific  rate  of  net  proceeds.  Reversely,  the  rate  of  brokerage  should  be  added 
to  any  obtained  rate  of  net  proceeds  to  find  a  required  quotation  or  rate  of 
gross  proceeds. 

NOTE  4.  The  product  of  two  factors  being  the  same  irrespective  of  the 
order  of  their  arrangement,  a  favorite  practise  among  brokers  is  to  place  the 
specific  rate  of  the  required  percentage  in  the  usual  position  of  the  multiplicand 
and  to  regard  it  as  the  required  dollars  per  share;  and  to  place  one-hundredth 
of  the  par  value  or  base  in  the  usual  position  of  the  multiplier,  and  to  regard 
it  as  the  number  of  shares. 

EXAMPLES  FOR  PRACTISE 

[In  the  following  examples,  the  par  value  per  share  is  understood  to  be 
$100,  when  not  otherwise  expressed.] 

Find  the 

1.  Dividend  on  $4700  par  of  stock,  rate  of  dividend  5  %. 
Par  value  or  base  X  specific  rate  of  dividend  =  dividend. 

2.  Assessment  on  $6200  par  of  stock,  rate  of  assessment  3?  %. 

3.  Dividend  on  163  shares  of  stock,  rate  of  dividend  7%. 

4.  Premium  on  $1900  par  of  stock,  rate  of  premium  4f  %. 

5.  Discount  on  72  shares  of  stock,  rate  of  discount  12f  %. 

Find  the  market  value  of 

6.  $2300  par  of  bonds  if  the  rate  of  premium  is  6J  %. 

Par  value  or  base  X  rate  of  market  value  [Note  1],  =  market  value. 

7.  59  shares  of  stock  if  the  rate  of  discount  is  16f  %. 

8.  $2700  par  of  bonds  if  the  rate  of  premium  is  12f  %. 

9.  148  shares  of  stock  if  the  quotation  is  87 J. 

10.  $3700  par  of  stock  if  quoted  at  18  f  %  discount. 

11.  $8600  par  of  bonds  if  quoted  at  29|  %  premium. 

Including  \  %  brokerage,  find  the  cost  of 

12.  $6800  par  of  stock  if  bought  at  9|  %  premium. 
Par  value  or  base  X  rate  of  gross  cost  [Note  2],  =  gross  cost. 

13.  43  shares  of  stock  if  bought  at  118f . 

14.  $9300  par  of  bonds  if  bought  at  23|  %  premium. 

15.  $8500  par  of  stock  if  bought  at  2J  %  discount. 


STOCKS   AND    BONDS  209 

Including  J  %  brokerage,  find  the  net  proceeds  of 

16.  $1800  par  of  bonds  if  sold  at  105f . 

Par  value  X  rate  of  net  proceeds  [Note  3],  =  net  proceeds. 

17.  37  shares  of  stock  if  sold  at  7f  %  discount. 

18.  $2500  par  of  bonds  if  sold  at  109J. 

19.  $1200  par  of  stock  if  sold  at  8f  %  premium. 

If  the  par  value  of  stock  is 

20.  $7300,  the  market  value  $7592,  find  the  %  of  premium. 
Percentage  of  market  value  -5-  par  value  or  base  =  rate  of  market  value. 

Rate  of  market  value  or  quotation  —  rate  of  par  (100  %)  =  specific  rate  of 
premium. 

21.  $2800,  market  value  $3433.50,  find  quotation  [205,  c]. 

22.  $5400,  the  market  value  $6277.50,  find  the  %  of  premium. 

23.  $8200,  market  value  $6211.50,  find  the  %  of  discount. 

24.  $3900,  premium  $755.63,  find  %  of  premium  [205,  a]. 
26.  $3600,  the  discount  $265.50,  find  the  %  of  discount. 

If,  allowing  J  %  brokerage,  the  cost  of  > 

26.  73  sh.  stock  is  $7637.63,  at  what  %  prem.  were  they  bought?   v 

Gross  cost  -5-  par  value  or  base  =  rate  of  gross  cost. 

Rate  of  gross  cost  —  rate  of  brok.  =  rate  of  prime  cost,  or  quo.  [Note.  2]. 

Quotation  —  rate  of  par  (100  %)  =  specific  rate  of  required  premium. 

27.  163  shares  is  $14996,  at  what  quotation  were  they  bought? 

28.  52  shares  is  $4329,  at  what  %  discount  were  they  bought? 

29.  $8400  par  of  stock  is  $9544.50,  at  what  %  premium  were 
they  bought? 

If,  allowing  \  %  brokerage,  the  net  proceeds  of 

30.  63  shares  are  $5213.25,  at  what  %  discount  were  they  sold? 
Net  proceeds  -H  par  value  or  base  =  rate  of  net  proceeds. 

Rate  of  net  pro.  +  rate  of  brok.  =  rate  of  gross  pro.,  or  quotation  [Note  3]. 
Rate  of  par  (100  %)  —  rate  of  quo.  =  specific  rate  of  required  discount. 

31.  $3400  par  of  stock  are  $3574.25,  at  what  quo.  was  it  sold? 

32.  92  sh.  stock  are  $7854.50,  at  what  %  disc,  were  they  sold? 

33.  126  shares  of  stock  are  $15576.75,  at  what  %  premium 
were  they  sold? 


210  .        PERCENTAGE 

Find  the  par  value,  if  the 

34.  Premium  is  $191.25,  and  the  rate  of  premium  12f  %. 
Percentage  of  prem.  -r-  specific  rate  of  prem.  =  par  value  or  base. 

35.  Discount  is  $954.75,  and  the  rate  of  discount  8f  %. 

36.  Dividend  is  $829.50,  and  the  rate  of  dividend  3J  %. 

37.  Assessment  is  $49.50,  and  the  rate  of  assessment  5J  %. 

38.  Market  value  is  $5875.25,  and  the  quotation  82f . 

39.  Market  value  is  $9634.25,  and  the  rate  of  premium  8i  %. 

40.  Market  value  is  $5372.25,  and  the  rate  of  discount  7|  %. 

Allowing  J  %  brokerage,  what  par  value  of  stock  can  be 

41.  Bought  for  $6424.75,  if  quoted  at  103J? 

Quo.  or  rate  of  prime  cost  +  rate  of  brok.  =  specific  rate  of  gross  cost. 
Percentage  of  gross  cost  -4-  its  specific  rate  =  its  par  value  or  base. 

42.  Bought  for  $2259.75,  if  the  quotation  is  98|? ' 

43.  Bought  for  $6243.75,  if  the  rate  of  premium  is  38f  %? 

44.  Bought  for  $2932,  if  the  rate  of  discount  is  8}  %? 

45.  Sold  for  $10631.25,  if  the  quotation  is  78|? 

Quo.  or  rate  of  gross  proceeds  —  rate  of  brok.  =  rate  of  net  proceeds. 
Percentage  of  net  proceeds  -5-  its  specific  rate  =  its  par  value  or  base. 

46.  Sold  for  $1384.50,  if  the  quotation  is  115J? 

47.  Sold  for  $5577,  if  the  premium  is  26J  %. 

48.  Sold  for  $8066.75,  if  the  discount  is  1J  %? 

/.^Jjv^?.   How  many  shares  of  stock  can  be  bought  at  2f  %  discount 
^      5nd  sold  at  If  %  premium,  to  net  a  profit  of    $297,  allowing  J  % 
>       brokerage  for  buying  and  the  same  for  selling? 

.j^X  50.   What  are  the  net  proceeds  of  126  shares  of  stock  sold  at 

1S|  %  discount,  brokerage  f  %? 

/     51.   A  gentleman  bought  stock  at  15i  %  premium,  brokerage 
\  %,  and  afterwards  drew  a  dividend  of  $480  upon  the  stock 
wjiich  he  purchased.     If  the  rate  of  dividend  was  5  %,  what  was 
yrfie  total  cost  of  the  stock? 

y         52.   A  man  bought  80  shares  of  stock  at  98 J ;  and  subsequently 
sold  36  shares  at  99 J,  and  the  remainder  at  99f .     Allowing  \  % 
brokerage  for  buying  and  for  selling,  what  was  his  net  gain? 
/      53.   I  receive  an  annual  income  of  $280  from  5  %  bonds  bought 
at  92f ,  brokerage  f  %.     What  was  the  cost  of  the  bonds? 


RELATION    OF    INCOME   TO    INVESTMENT  211 

54.  A  man  invested  $19857.50  in  4%  bonds  purchased  at 
84f ,  brokerage  £  %.     What  was  his  annual  income  from  the 
purchase? 

55.  An  investor  derives  a  quarterly  income  of  $525  from  6  % 
bonds  bought  at  28J%  discount,  brokerage  \%.     What  did  he 
pay  for  the  bonds? 

56.  If  $18716.25  is  realized  as  net  proceeds  from  the  sale  of 
13800  par  of  bonds,  at  what  %  premium  were  they  sold,  allowing 

\  %  for  brokerage? 

RELATION    OF    INCOME   TO    INVESTMENT 

273.  INTRODUCTION,  (a)  As  viewed  by  a  corporation  when  it 
proposes  to  distribute  a  dividend  or  to  levy  an  assessment,  etc., 
the  par  or  face  value  of  the  stock  would  be  the  only  unchanging 
base  that  could  be  used  as  a  standard  in  prorating  such  a  divi- 
dend or  assessment  among  its  stockholders  which  would  apply 
generally  to  all  its  shares,  irrespective  of  their  constantly  fluctu- 
ating market  value  on  the  different  days  of  their  purchase  by  the 
present  holders  of  the  stock.  From  the  standpoint  of  the  inves- 
tor, however,  an  entirely  different  standard  would  suggest  itself 
in  considering  the  expediency  of  an  investment  in  a  particular 
class  of  securities.  The  thought  uppermost  in  his  mind  would 
very  naturally  be  the  %  of  income  which  he  would  realize  upon  his 
investment,  rather  than  the  %  of  income  he  would  realize  upon 
the  par  value.  Other  things  being  equal,  a  high  rate  of  dividend 
on  stocks  or  of  interest  on  bonds,  the  market  value  of  which  is 
at  a  high  premium,  may  not  be  so  desirable  as  a  lower  rate  of  divi- 
dend on  stocks  or  of  interest  on  bonds  the  market  value  of  which 
is  at  a  lower  premium,  or  possibly  at  a  discount.  Hence,  in  con- 
sidering the  expediency  of  a  particular  purchase,  investors  regard 
the  sum  invested  (gross  cost)  as  the  base. 

(b)  If  the  investment  is  the  contemplated  base,  it  should  be 
expressed  by  placing  the  phrase  "on  investment"  immediately 
after  the  rate  %  which  refers  to  such  a  base.  Otherwise,  in  many 
instances,  the  par  value  may  be  misunderstood  as  the  intended 
base.  Thus,  in  the  expressions  "5%  stocks,"  "6%  bonds,"  etc., 
no  base  being  expressed,  they  are  respectively  understood  to  refer 
to  stocks  paving  5%  dividends  on  par  value,  bonds  paying  6% 
interest  on  par  value;  but  when  reference  is  made  to  "  stock  pay- 
ing 5  %  income  on  investment,"  it  is  meant  that  the  stock  referred 
to  pays  a  dividend  on  its  par  value  which  is  equivalent  to  5  %  on 
the  necessary  investment  to  purchase  it. 


212  PERCENTAGE 

(c)  If,  in  the  same  relation,  two  rates  %  of  income  are  men- 
tioned, one  of  which  is  understood  to  be  on  par  value  and  the 
other  is  expressed  on  investment,  the  rate  %  of  income  on  par 
(100)  may  be  regarded  as  that  many  dollars  of  income  per  share 
(a  percentage  or  product),  and  the  rate  %  of  income  on  invest- 
ment may  be  retained  as  the  specific  rate  of  that  income,  or 
multiplier,  (d)  Similarly,  any  rate  %  on  par  value  is  convert- 
ible into  the  same  number  of  dollars  per  share  of  the  same  specific 
name  as  the  rate,  (e)  Conversely,  any  result  obtained  in  dollars 
per  share  may  be  accepted  as  so  much  %  on  par  value  of  the  same 
specific  name. 

Identification  of  terms.  1.  The  indicated  investment  is  the 
base  or  multiplicand.  2.  The  %  of  income  on  investment  is  the 
rate  or  multiplier.  3.  The  income  is  the  percentage  or  product. 

EXAMPLES  FOR  PRACTISE 

At  what  price  per  share  must 

'..  7  %  bonds  be  bought  to  realize  8  %  income  on  investment? 
7  %  bonds  are  bonds  paying  $7  interest  on  $100  [Intro.,  c].     Hence, 
Percentage  of  income  ($7)  •+•  its  specific  rate  on  inv.  (8  %)  =  $87|,  inv. 

.  6  %  stock  be  bought  to  realize  10  %  income  on  investment? 

.  5  %  bonds  be  bought  to  produce  4  %  income  on  investment? 

*/4.  8  %  stock  be  bought  to  yield  5  %  income  on  investment? 

l/tfi  9  %  bonds  be  bought  to  produce  8  %  income  on  investment? 

Including  f  %  brokerage,  at  what  'quotation  must 
^/'  6.   5  %  stock  be  bought  to  yield  8  %  income  on  investment? 

5  %  stock  means  stock  paying  $5  dividend  per  $100  share.  Hence,  inc. 
per  sh.  ($5)  -f-  rate  of  inc.  on  inv.  (8  %)  =  inv.  per  sh.  ($62.50).  Investment 
or  gross  cost  per  share  ($62|)  —  brokerage  per  share  ($|)  =  prime  cost  per  sh. 
($62 f ),  or  rate  of  quotation  (62|)  [See  Introduction,  e]. 

»/7.  4  %  bonds  be  bought  to  realize  5  %  income  on  investment? 

8.  6  %  stock  be  bought  to  yield  8%  income  on  investment? 

,9.  7%  bonds  be  bought  to  yield  10%  inc.  on  investment? 

^lO.  3  %  stock  be  bought  to  yield  4  %  income  on  investment? 

What  %  of  income  on  investment  will  be  realized  if 
/ll.   4  %  stock  be  bought  at  80. 

4  %  stock  means  stock  paying  $4  dividend  on  each  $100  share.  As  there 
is  no  brokerage,  the  quotation  or  prime  cost  per  share  also  expresses  the 


RELATION    OF   INCOME   TO   INVESTMENT  213 

investment  or  gross  cost  per  share  ($80).     Hence,  income  per  share  ($4)  -s- 
in vestment  per  share  ($80)  =  specific  rate  of  inc.  on  inv.  (5  %). 

i/12.  6  %  stock  is  bought  at  75? 

^  13.  5  %  bonds  are  bought  at  25  %  premium? 

14.  4J  %  stock  is  bought  at  10  %  discount? 

15.  4  %  bonds  are  bought  at  20  %  discount? 

8  %  stock  be  bought  at  119J  %,  brokerage  J? 
Income  per  sh.  ($8)  -f-  inv.  per  sh.  ($119£  -f  $|)  =  specific  rate  of  income 
on  the  investment  base  (6|)  %. 

3  %  bonds  be  bought  at  74  J,  brokerage  i  %? 
18.  4  %  stock  be  bought  at  89 J,  brokerage  \  %? 

5  %  bonds  be  bought  at  79 J,  brokerage  J  %? 
'20.  4£  %  stock  be  bought  at  112f ,  brokerage  \  %? 

WhaJ/%  stock  can  be  bought 

1/21.   At  75,  and  yield  4  %  income  on  investment? 

Inv.  per  sh.  or  base  ($75)  X  rate   %  of  inc.  on  that  inv.  (4  %)  =  income 
per  sh.  from  that  investment  ($3),  or  3  %  stock  [See  Introduction,  e]. 

f/22.  At  120,  and  produce  5  %  income  on  investment? 

23.  At  25  %  premium  and  yield  8  %  income  on  investment? 

24.  At  20  %  discount  and  realize  10  %  income  on  hi  vestment? 

25.  At  124J  and  yield  4  %  on  investment,  brokerage  J  %? 

^r     Investment  per  share  or  base  ($124 f  +  $}  =  $125)  X  rate   %  of  income 
on  that  investment  (4  %)  =  income  per  share  ($5),  or  5  %  stock. 

26.  At  74|  and  yield  8  %  income  on  investment,  brokerage  \  %? 

27.  At  79f  and  produce  5  %  income  on  investment,  brok.  \  %? 

28.  At  112f  and  realize  4  %  income  on  investment,  brok.  \  %? 

29.  What  %  on  investment  is  realized  in  purchasing  a  house 
and  lot  for  $4000  which  can  be  rented  at  $50  per  month,  allowing 
$100  per  annum  for  deterioration  in  value  and  $180  per  annum 
for  taxes,  repairs  and  other  expenses? 

30.  What  %  on  investment  is  realized  in  buying  9  %  stock  at 
74|,  allowing  \  %  for  brokerage? 

31.  What  %  bonds  can  be  bought  at  87f  to  yield  an  income  of 
8  %  on  investment,  brokerage  \  %? 

32.  At  what  quotation  must  3  %  bonds  be  purchased  to  pro- 
duce an  income  of  5  %  on  investment,  brokerage  \  %? 


214  PERCENTAGE 

33.  An  investor  paid  $6580  for  stock  from  which  he  derived 
an  annual  income  of  $329.     What  %  of  income  upon  his  invest- 
ment did  he  realize? 

34.  Both  being  equally  secure,  which  is  the  more  profitable 
investment,  5%  stock  bought  at  20%  premium,  or  4%  stock 
bought  at  10  %  discount,  and  how  much  better? 

35.  A  man  bought  74  shares  of  stock  at  85f ,  and  afterwards 
sold  the  same  at  86 J.     What  was  his  gain,  allowing  £  %  brokerage 
for  buying  and  selling? 

36.  An  investor  bought  50  shares  of  stock  at  If  %  premium, 
drew  a  dividend  thereon  of  4  %,  and  sold  it  at  98|.     What  was  his 
gain,  allowing  f  %  brokerage  each  way? 

37.  A  man  bought  300  shares  of  stock  at  109|,  sold  90  shares 
at  110J,  150  shares  at  11  Of,  and  the  remainder  at  11  If.     Allowing 
i  %  for  brokerage  in  each  transaction,  what  was  his  net  gain? 


EXCHANGE 

274.  INTRODUCTION,  (a)  Exchange  is  the  giving  of  one  thing 
as  an  equivalent  for  another.  Therefore,  when  a  credit  in  one 
city  where  it  is  not  needed  is  bartered  for  a  credit  in  another  city 
where  it  is  required,  at  a  less  cost  than  the  expressage  for  safely 
transmitting  the  credit  money  from  the  one  city  to  the  other, 
the  transaction  is  very  appropriately  termed  " exchange."  Thus, 
if  A  of  Baltimore  owes  $5000  to  B  of  New  Orleans;  and  C  of 
New  Orleans  owes  $5000  to  D  of  Baltimore,  and  this  was  known 
to  the  parties  concerned,  the  simplest  way  to  dispose  of  both  debts 
would  evidently  be  for  A  in  Baltimore  to  pay  $5000  to  D  of  the 
same  city,  and  for  C  of  New  Orleans  to  pay  $5000  to  B  of  that 
city;  for  this  exchange  of  credits  would  involve  no  expressage  for 
transmission  of  money  from  either  place  to  the  other.  Any  such 
arrangement,  however,  is  rendered  impracticable  by  the  unlike- 
lihood of  the  persons  named  possessing  the  requisite  knowledge 
to  make  such  an  exchange  feasible.  To  make  such  a  convenient 
disposition  of  credits  effective,  " exchange  banks"  are  established 
in  the  principal  commercial  cities  of  the  world,  each  exchange 
bank  in  one  city  having  as  its  correspondent  a  similar  bank  in 
other  cities,  through  which  it  can  transact  its  outside  business. 
Thus,  an  exchange  bank  in  Baltimore  can  cash  D's  draft  on  C,  as 
well  as  the  drafts  of  other  Baltimore  firms  on  New  Orleans,  and 


EXCHANGE  215 

mail  them  for  collection  to  its  correspondent  bank  in  New  Orleans, 
the  proceeds  to  be  held  in  New  Orleans  subject  to  the  order  of 
the  Baltimore  bank;  and  A  of  Baltimore  would  very  naturally  go 
to  his  local  bank  to  purchase  a  draft  of  $5000  upon  its  New  Orleans 
correspondent  and  mail  it  to  B,  discharging  in  this  manner  his 
indebtedness  without  the  cost  of  transmitting  the  actual  money, 
while  the  Baltimore  bank  is  also  saved  the  expense  of  expressing 
from  New  Orleans  to  Baltimore  the  proceeds  of  D's  draft  on  C  by 
utilizing  it  in  paying  A's  debt  in  that  city  to  B,  besides  receiving 
a  fee  for  cashing  D's  draft  on  C,  and  a  fee  for  selling  to  A  a  draft 
on  its  New  Orleans  correspondent  in  favor  of  B. 

(6)  If  New  Orleans  owes  Baltimore  about  as  much  as  Balti- 
more owes  New  Orleans  the  course  of  exchange  between  the  two 
cities  is  said  to  be  at  par  (that  is,  on  an  equality) ;  but  if  New  Or- 
leans owes  Baltimore  more  than  Baltimore  owes  New  Orleans, 
the  course  of  exchange  in  New  Orleans  will  be  at  a  premium  on 
Baltimore,  the  amount  of  the  premium  varying  with  the  amount 
of  the  preponderance  of  indebtedness  and  the  consequent  cost  of 
transmitting  the  actual  money  to  overcome  the  preponderance; 
and  at  the  same  time  the  course  of  exchange  in  Baltimore  on  New 
Orleans  will  be  at  a  discount  to  stimulate  the  buying  of  exchange 
on  New  Orleans  and  thus  to  diminish  the  preponderance.  The 
highest  premium  or  lowest  discount  can  be  but  little  more  than 
the  cost  of  safely  transmitting  the  sum  of  money  mentioned  on 
the  face  of  the  bill  of  exchange  from  the  debtor  city  to  the  cred- 
itor city. 

275.  A  banker's  draft  or  bill  of  exchange  is  an  order  written 
by  one  bank  (called  the  drawer  of  the  exchange),  directing  another 
bank  (called  the  drawee  of  the  exchange),  to  pay  a  specified  sum 
of  money  to  a  designated  person  (called  the  payee)  or  to  whomso- 
ever the  payee  orders  it  to  be  paid. 

NOTE.  When  the  drawer  and  drawee  reside  in  the  same  country,  the  draft 
is  called  a  domestic  bill  of  exchange;  and  when  they  reside  in  different  coun- 
tries, it  is  named  a  foreign  bill  of  exchange. 

276.  The  face  or  par  of  a  bill  of  exchange  is  the  sum  specified 
therein  to  be  paid  to  the  payee  or  to  his  order. 

NOTE.  If,  as  is  usually  the  case,  a  debt  is  payable  where  the  payee 
resides,  the  face  of  a  bill  of  exchange  in  payment  of  that  debt  should  be  the 
sum  due  to  the  payee  by  the  purchaser  of  the  bill,  the  purchaser  defraying  the 
necessary  premium  charged,  or  receiving  the  benefit  of  any  discount 
allowed. 


216  PERCENTAGE 

DOMESTIC    EXCHANGE 

277.  Domestic  exchange  is  exchange  effected  between  cities  of 
the  same  country. 

NOTE.  The  method  of  computing  domestic  exchange  is  applicable  to  all 
countries  which  use  the  same  coinage,  and  which  must  therefore  express  the 
rate  of  exchange  directly  by  premium  or  discount  on  the  one  common  coinage. 

278.  Identification  of  terms.     1.    The  face  of  a  bill  of  exchange 
is  the  base  or  multiplicand.     2.    The  %  of  any  required  premium, 
discount,  or  market  value,  is  a  rate  or  multiplier.     3.    The  premium, 
the  discount,  or  the  market  value,  is  a  percentage  or  product. 

NOTE  1.  The  rate  of  market  value  of  a  sight  draft  equals  the  rate  of  its 
face  value  (100  %)  increased  by  the  rate  %  of  premium,  or  diminished  by  the 
rate  %  of  discount. 

NOTE  2.  The  rate  of  market  value  of  a  time  draft  equals  the  rate  of 
market  value  of  a  sight  draft  [Note  1]  diminished  by  the  specific  rate  of  in- 
terest for  the  given  time  [Note,  227].  Hence,  if  both  are  of  the  same  face 
value  and  bought  at  the  same  rate  of  exchange,  the  cost  of  a  time  draft  is  as 
much  less  than  that  of  a  sight  draft,  as  will  equal  the  interest  on  its  face  (the 
bank  discount)  for  the  given  time. 

NOTE  3.  As  drafts  are  bought  and  sold  at  their  market  value,  their 
market  value  must  also  be  their  prime  cost  if  bought,  or  their  gross  proceeds 
if  sold.  Hence,  if  brokerage  is  included,  it  should  be  computed  upon  the 
market  value  or  prime  cost,  and  added  to  the  prime  cost  to  find  the  gross  cost, 
if  purchased;  or  computed  upon  the  market  value  or  gross  proceeds,  and  sub- 
tracted from  the  gross  proceeds  to  find  the  net  proceeds,  if  sold. 

NOTE  4.  (a)  If  a  debt  is  payable  at  the  place  in  which  the  debtor  resides, 
the  amount  of  the  debt  should  be  regarded  as  the  gross  cost  of  the  draft,  that  is, 
the  creditor  should  assume  the  expense  of  remittance;  (6)  but  if  the  debt  is 
payable  at  the  place  in  which  the  creditor  resides,  the  amount  of  the  debt 
should  be  taken  as  the  face  of  the  draft,  that  is,  the  purchaser  of  the  draft 
(the  debtor)  is  properly  chargeable  with  the  expense  of  remittance. 

EXAMPLES  FOR  PRACTISE 

1.  Find  the  cost  of  a  sight  draft  of  $720  at  f  %  premium. 
Face  of  draft  or  base  ($720)  X  rate  or  prem.  (f  %)  =  premium  ($2.70). 
Face  of  draft  ($720)  +  prem.  ($2.70)  =  market  value  or  cost  ($722.70). 

2.  Find  cost  of  a  sight  draft  of  $630  at  J  %  premium. 

3.  Find  market  value  of  a  sight  draft  of  $960  at  f  %  discount. 


DOMESTIC    EXCHANGE  217 

4.   Find  proceeds  of  a  sight  draft  of  $568.30  at  J  %  premium. 
6.   Find  cost  of  a  sight  draft  of  $275.85  at  J  %  discount. 

6.  Find  cost  of  a  30-day  draft  of  $328;  prem.  f  %,  interest  6  %. 
Face  of  draft  or  base  ($328)  X  rate  or  prem.  (f  %)  =  premium  ($2.05). 
Interest  of  $328  for  30  days  at  6  %  =  $1.64,  discount  for  interest. 

Face  ($328)  +  prem.  ($2.05)  -  int.  ($1.64)  =  cost,  $328.41  [Note  4,  a]. 

7.  Find  cost  of  a  60-day  draft  of  $843.75;  prem.  \  %,  int.  6  %. 

8.  Find  cost  of  a  90-day  draft  of  $476.28;  prem.  f  %,  int.  5  %. 

9.  Find  pro.  of  a  30-day  draft  of  $624;  prem.  f  %,  int.  4%. 

10.  Find  cost  of  a  60-day  draft  of  $460;  disc.  \  %,  int.  6%. 
Face  of  draft  ($460)  X  rate  of  disc.  (\  %)  =  discount  ($1.15). 
Interest  of  $460  for  60  days  at  6  %  =  $4.60,  discount  for  interest. 

Face  of  draft  ($460)  -  disc.  ($1.15)  -  int.  ($4.60)  =  cost  ($454.25). 

11.  Find  cost  of  90-day  draft  of  $728.16;    disc.  J%,  int.  5%. 

12.  Find  pro.  of  30-day  draft  of  $640;   disc,  f  %,  int.  6%. 

13.  Find  market  value  of  60-day  draft  of  $160;  discount  |  %, 
interest  6%. 

Find  the  face  of  a  sight  draft  which  can  be  bought 

14.  For  $1239.09,  if  the  rate  of  exchange  is  £  %  premium. 
Percentage  of  market  value  or  product  ($1239.09)  -5-  the  specific  rate  of 

market  value  or  multiplier  [100  %  +  J  %,  Note  1]  =  face  value  or  multi- 
plicand, $1236.  [Note  4,  b]. 

16.  For  $875.27,  if  the  rate  of  exchange  is  f  %  premium. 

16.  For  $932.85,  if  the  rate  of  exchange  is  f  %  discount. 

17.  For  $1465.20,  if  the  rate  of  exchange  is  J  %  premium. 

18.  For  $716.50,  if  the  rate  of  exchange  is  J  %  discount. 

Find  the  face  of  a  time  draft  which  can  be  bought  for 

19.  $824.25,  payable  in  90  days;   exch.  f  %  disc.,  int.  6%. 
Rate  of  face  (100  %)  —  rate  of  exchange  discount  (f  %  =  .00375)  —  rate 

of  int.  for  90  da.  at  6  %  per  annum  (.015)  =  rate  of  market  value  (.98125). 
Percentage  of  market  value  or  product  ($824.25)  -f-  specific  rate  of  market 
value  or  multiplier  (.98125)  =  face  value  or  multiplicand  ($840). 

20.  $753.25,  if  payable  in  30  days,  exch.  J%  prem.,  int.  5%. 

21.  $1581.70,  if  payable  in  60  days,  exch.  f  %  disc.,  int.  4J  %. 

22.  $3500,  if  payable  in  90  days,  exch.  \  %  prem.,  int.  6  %. 


218  PERCENTAGE 

23.  $875,  if  payable  in  60  days,  exch.  J  %  disc.,  int.  4%. 

24.  The  net  proceeds  of  a  sale  were  $2785.90,  which  the  com- 
mission merchant  remitted  to  his  principal  by  a  bank  draft,  pay- 
able at  sight,  purchased  at  f  %  discount.    What  was  the  face  of  the 
draft?    [Note  4,  a] 

25.  A  St.  Louis  merchant  bought  $3258.25  worth  of  goods 
from  a  firm  in  New  York  payable  in  60  days,  and  when  the  bill 
became  due  he  remitted  a  bank  draft  to  New  York  in  settlement. 
What  was  the  cost  of  the  draft,  if  the  course  of  St.  Louis  exchange 
on  New  York  was  at  J  %  premium?   [Note  4,  b] 

FOREIGN    EXCHANGE 

279.  Foreign  exchange  is  exchange  between  different  countries 
which  also  have  different  monetary  units. 

280.  The  rate  of  foreign  exchange  is  the  market  value  of  the 
monetary  unit  of  the  country  in  which  the  drawee  resides  when  it 
is  estimated  in  the  currency  of  the  country  in  which  the  bill  of 
exchange  is  purchased. 

NOTE  1.  The  intrinsic  par  of  exchange  is  the  inherent  or  true  value  of 
such  foreign  coins  as  constitute  the  monetary  units  of  their  respective  coun- 
tries, as  shown  by  the  weight  and  purity  of  the  gold  or  silver  which  they 
contain.  In  138  will  be  found  a  TABLE  showing  the  intrinsic  par  of  the  prin- 
cipal foreign  monetary  units  when  expressed  in  U.  S.  money,  as  proclaimed  by 
the  Secretary  of  the  Treasury  of  the  United  States  on  January  1,  1912. 

NOTE  2.  The  commercial  par  of  exchange,  or  the  rate  of  foreign  exchange, 
is  the  market  value  of  foreign  monetary  units  as  determined  by  the  balance 
of  trade  between  the  country  of  the  drawer  and  that  of  the  drawee  [274,  b]. 
If  the  balance  of  trade  is  against  the  country  on  which  the  bill  of  exchange  is 
drawn,  that  is,  if  it  has  bought  more  from  the  country  of  the  drawer  than  it 
has  sold  to  it,  the  commercial  par  of  exchange  on  the  country  of  the  drawee 
will  be  at  a  discount  upon  the  intrinsic  par;  but  if  the  balance  of  trade  is  in 
favor  of  the  country  on  which  the  draft  is  drawn,  the  commercial  par  will  be 
quoted  at  a  premium  upon  the  intrinsic  par  [274,  b]. 

NOTE  3.  In  the  United  States,  the  rate  of  exchange  on  Great  Britain  is 
expressed  by  giving  the  market  value  of  £1  in  United  States  money;  on  France, 
Belgium  and  Switzerland  by  giving  the  market  value  of  $1  in  francs;  on  Ger- 
many by  giving  the  market  value  of  4  reichsmarks  in  United  States  money; 
and  on  Holland  by  giving  the  market  value  of  1  guilder  in  United  States 
money.  Hence, 


FOREIGN    EXCHANGE  219 

IDENTIFICATION   OF  TERMS 

281.  If  the  rate   of   exchange    expresses  the  market  value  of 
1  foreign   monetary   unit   in   United   States  money.       1.    The  ex- 
change value  in  United  States  money  of  1  foreign  monetary  unit  is 
the  multiplicand.     2.    The  number  of  foreign  monetary  units  is  the 
multiplier.      3.    The  exchange  value  in  United  States  money  of  all 
the  foreign  monetary  units  is  the  product. 

282.  If  the  rate  of  exchange  expresses  the  market  value  of  $i 
in  foreign  monetary  units.     1.    The  number  of  foreign  monetary 
units  that  can  be  bought  for  $1  is  the  multiplicand.     2.    The  number 
of  dollars  to  be  invested  in  exchange  is  the  multiplier.     3.    The  num- 
ber of  foreign  monetary  units  that  can  be  bought  with  all  the  dollars 
to  be  invested  in  exchange  is  the  product. 

NOTE  1.  As  American  exchange  upon  Germany  expresses  the  value  in 
United  States  money  of  4  German  reichsmarks,  the  rate  of  exchange  must  be 
divided  by  4  to  obtain  the  multiplicand  of  281. 

NOTE  2.  Notes  2,  3  and  4,  278,  apply  also  to  foreign  exchange,  except 
that  in  computing  interest  upon  time  bills  of  exchange  on  Great  Britain, 
3  days  of  grace  are  invariably  added  to  the  nominal  time.  Grace  is  also 
allowed  in  a  few  American  cities  upon  domestic  time  bills  of  exchange. 

TO  FIND  THE  COST  OF  A  FOREIGN  BILL  OF  EXCHANGE 

283.  The  following  forms  of  solution  are  employed  when  the 
buyer  of  the  exchange  owes  money  which  is  payable  where  the 
payee  resides.     The  face  of  the  bill  should  then  be  the  amount  of 
the  debt,  and  any  expense  incurred  in  buying  the  bill  of  exchange 
should  be  borne  by  the  remitter. 

ILLUSTRATIVE  EXAMPLES 

1.  What  is  the  cost  of  a  bill  of  exchange  on  London  for  £673  2s. 
6d.,  if  the  rate  of  exchange  is  4.8725? 

SOLUTION.  £673  2s.  6d.  =  £673.125  [111.  Ex.  2,  160].  $4.8725  is  the 
exchange  value  in  United  States  money  of  £1,  or  the  multiplicand  [281,  1]; 
£673.125  is  the  number  of  £'s,  or  the  multiplier  [281,  2];  therefore,  multiply 
these  factors  ($4.8725  X  673.125  =  $3279.80)  to  find  the  exchange  value  in 
United  States  money  of  all  the  given  £'s,  or  product  [281,  3]. 

2.  What  is  the  cost  of  a  bill  of  exchange  on  Berlin  for  675.20 
reichsmarks  if  the  rate  of  exchange  is  96 J? 


220  PERCENTAGE 

SOLUTION.  $.96£  4-  4  =  $.24i  [Note.  1,  282],  the  exchange  value  of  1 
reichsmark,  or  the  multiplicand  [281,  1];  675.20  is  the  number  of  reichsmarks, 
or  the  multiplier  [281,  2].  Therefore  multiply  these  factors  ($.24i  X  675.20 
=  $162.89  -|-)  to  find  the  exchange  value  of  all  the  reichsmarks,  or  the  product 
[281,  3]. 

3.  What  is  the  cost  of  a  bill  of  exchange  on  Paris  for  3245.80 
francs,  if  the  rate  of  exchange  is  5.20$? 

SOLUTION.  5.20j  is  the  number  of  francs  that  can  be  bought  for  $1,  or 
the  multiplicand  [282,  1];  and  3245.80  is  the  number  of  francs  to  be  bought 
with  all  the  required  dollars  to  be  invested,  or  the  product  [282,  3].  There- 
fore divide  the  given  product  (3245.80  fr.)  by  its  given  multiplicand  factor 
(5.20i  fr.)  to  find  its  required  multiplier  factor  (623.89  times  $1,  or  $623.89), 
the  number  of  dollars  to  be  invested. 

EXAMPLES  FOR  PRACTISE 

Find  the  cost  of  a  bill  of  exchange  for 

1.  £416  10s.,  if  the  rate  of  exchange  is  4.8650. 

2.  £725  17s.  6d.,  the  rate  of  exchange  being  4.8715. 

3.  £926  12s.  6d.,  if  the  rate  of  exchange  is  4.8620. 

4.  827.60  reichsmarks,  if  rate  of  exch.  is  96  [Note  1,  282]. 

5.  1927.50  reichsmarks,  if  the  rate  of  exchange  is  95f . 

6.  3460.25  reichsmarks,  if  the  rate  of  exchange  is  96£. 

7.  4675.20  francs,  if  the  course  of  exchange  is  5.18|. 

8.  798.40  francs,  the  rate  of  exchange  being  5.19J. 

9.  1425.80  francs,  if  the  rate  of  exchange  is  5.18|. 

10.  2300  guilders,  if  the  course  of  exchange  is  40f. 

11.  £826  2s.  6d.;  exch.  4.8665,  brok.  J%  [Note  3,  278]. 

12.  6350  francs,  if  exchange  is  5.18J,  and  brokerage  J  %. 

TO  FIND  THE  FACE  OF  A  BILL  OF  EXCHANGE 

284.  The  following  forms  of  solution  are  applicable  when  the 
buyer  of  the  exchange  owes  money  which  is  payable  where  he 
(the  buyer)  resides.  The  total  cost  of  the  bill  of  exchange  should 
then  be  equal  to  the  amount  of  the  debt,  thus  causing  all  expenses 
incurred  in  buying  the  bill  to  be  borne  by  the  payee. 

ILLUSTRATIVE  EXAMPLES 

1.  I  paid  $925.70  for  a  bill  of  exchange  on  Liverpool.  What 
was  the  face  of  the  bill,  the  rate  of  exchange  being  4.86? 


BANKRUPTCY  221 

SOLUTION.  $925.70  is  the  exchange  value  of  all  the  required  £'s,  or  the 
product  [281,  3];  $4.86  is  the  exchange  value  of  £1,  or  the  multiplicand 
[281,  lj.  Therefore  divide  the  given  product  ($925.70)  by  its  given  multipli- 
cand factor  ($4.86)  to  find  the  required  number  of  £'s,  or  the  multiplier  factor 
(£190  9s.  51  d.). 

2.  I  bought  through  a  broker  a  bill  of  exchange  on  Belgium  for 
which  I  paid  him  $640.80.  If  the  rate  of  exchange  was  5.18,  and 
the  brokerage  |  %,  what  was  the  face  of  the  bill? 

SOLUTION.  $640.80  -s-  l.OOi  =  $640  [Note  3,  278],  the  number  of  dollars 
to  be  invested  in  exchange,  or  the  multiplier  [282,  2];  5.18  is  the  number  of 
francs  that  can  be  bought  for  $1,  or  the  multiplicand  [282,  1].  Therefore 
multiply  these  two  factors  (5.18  fr.  X  640  =  3315.20  fr.)  to  find  the  required 
number  of  francs  which  can  be  bought  with  all  the  dollars  to  be  invested  in 
the  bill  of  exchange,  or  the  product  [282,  3]. 

NOTE.  If  brokerage  is  included,  regard  the  amount  of  the  debt  to  be 
canceled  by  the  remittance  as  the  gross  cost  of  the  bill  of  exchange,  and 
divide  it  by  100  %  +  the  rate  of  brokerage  to  find  the  prime  cost  or  market 
value  of  the  bill  [Note  3,  278]. 

EXAMPLES  FOR  PRACTISE 

Find  the  face  of  a  bill  which  can  be  bought  for 

1.  $1825.36,  if  the  rate  of  exchange  is  4.8668  to  the  £. 

2.  $746.25,  if  the  rate  of  exchange  is  4.855  to  the  £. 

3.  $825.70,  if  the  course  of  exchange  is  5.19J  francs  to  $1. 

4.  $2580.20,  exchange  being  5.18J  francs  to  $1. 

5.  $942.65,  if  exchange  is  94f  £  to  4  reichsmarks. 

6.  $6250.75,  if  exchange  is  95i  $.  to  4  reichsmarks. 

7.  $675.80,  if  exchange  is  41  \i  to  1  guilder. 

8.  $1248.35,  if  exchange  is  40|^f  to  1  guilder. 

9.  $3650.25,  exchange  4.8575  to  £1,  brokerage  J  %. 
10.  $785.90,  exchange  41f  ?f  to  1  guilder,  brokerage  J%. 

BANKRUPTCY 

285.  INTRODUCTION.  A  firm  is  said  to  be  bankrupt  or  insol- 
vent when  its  financial  resources  are  not  sufficient  to  pay  its  debts 
as  they  fall  due.  The  assets  of  a  bankrupt  firm  are  its  total 
available  property,  including  the  collectible  debts  due  to  it  from 
others;  and  the  liabilities  of  a  bankrupt  firm  are  the  debts  it 
owes  to  others.  The  assignee  or  trustee  is  one  who  is  appointed 
to  take  charge  of  the  business  of  a  bankrupt  firm  for  the  purpose 


222  PERCENTAGE 

of  converting  its  assets  into  cash  sufficient  to  discharge  its  financial 
obligations;  or,  if  its  total  assets  are  insufficient  to  discharge  its 
total  liabilities,  to  distribute  among  its  creditors  such  a  percentage 
of  their  individual  claims  as  the  firm's  total  assets  when  con- 
verted into  cash  are  of  its  total  liabilities.  This  distribution  of 
available  assets  by  the  assignee  among  the  creditors  is  called  a 
dividend. 

Identification  of  terms.  1.  The  total  liability  of  the  firm  is  the 
base  or  multiplicand.  2.  The  %  of  the  total  liability  which  is 
distributed  among  a  bankrupt  firm's  creditors,  is  the  rate  of  dividend 
or  multiplier.  3.  The  sum  distributed  among  the  bankrupt  firm's 
creditors  is  the  percentage  of  dividend  or  product. 

NOTE  1.  The  total  dividend  to  all  the  creditors  will  always  be  the  net 
available  assets,  that  is,  the  gross  available  assets  diminished  by  the  expenses 
of  assignment. 

NOTE  2.  Whatever  the  rate  of  dividend  to  all  the  creditors  collectively 
may  be,  must  also  be  the  rate  of  dividend  to  each  individual  creditor.  Hence, 
the  total  liability  to  each  individual  creditor  is  a  base  or  multiplicand;  the  % 
of  dividend  is  the  rate  or  multiplier;  and  the  dividend  received  by  each 
individual  creditor  is  a  percentage  or  product. 

EXAMPLES 

1.  The  following  statement  was  rendered  by  the  assignee  of  a 
bankrupt   merchant:    Assets  —  merchandise   as   per   inventory, 
$8411.15;   personal  accounts  due  the  firm,  $1618.45;   real  estate, 
$4500.     Liabilities  —  notes  due  to  others,  $12000;   due  to  F.  A. 
Sadler  for  merchandise,  $6500;   due  to  R.  M.  Browning  for  ser- 
vices rendered,  $2700.     If  the  expenses  of  assignment  were  $325.60, 
what  dividend  should  F.  A.  Sadler  and  R.  M.  Browning  receive? 

$8411.15  +  $1618.45  +  $4500  =  $14529.60,  total  assets.  Total  assets 
($14529.60)  —  expenses  of  assignment  ($325.60)  =  net  assets  ($14204). 
$12000  +  $6500  +  $2700  =  $21200,  total  liabilities.  Percentage  of  total  net 
assets  or  product  ($14204)  -?-  total  liability,  the  base  or  multiplicand  ($21200) 
=  67  %,  the  rate  of  dividend  or  multiplier.  Hence, 

$6500  X  .67  =  $4355,  dividend  to  F.  A.  Sadler  [Note  2]. 

$2700  X  .67  =  $1809,  dividend  to  R.  M.  Browning. 

2.  A  firm  failed  with  liabilities  amounting  to  $33864.25  and 
assets  amounting  to  $24293.09.     What    %  of  dividend  will  the 
assignee  be  able  to  declare;    and  what  dividend  should  Geo.  C. 


INSURANCE  223 

Round  &  Co.  receive  whose  claim  for  $837.50  was  allowed,  if  the 
expenses  of  the  assignment  were  $1265.40? 

3.  The  available  resources  of  a  bankrupt  firm  were  $10284.75, 
and  its  liabilities  $12537.50.     If  the  expenses  of  assignment  were 
$756.25,  how  much  should  J.  P.  Smith  &  Co.  receive  whose  claim 
for  $1785.20  was  accepted? 

4.  A  bankrupt  firm  owed  A  $580,  B  $1275.50,  and  C  $875.30. 
Its    realizable    assets  were  $12142.25,  and    its    total  liabilities 
$17856.25.     If  the  expenses  of  settling  the  bankruptcy  were  4  % 
of  the  assets,  how  much  should  A,  B,  and  C  receive? 


INSURANCE 

286.  INTRODUCTION.  Insurance  is  a  pecuniary  indemnity 
guaranteed  by  one  party  (the  insurer)  to  another  party  (the  in- 
sured) in  certain  specified  contingencies,  such  as  the  death  of  the 
insured  (life  insurance)',  or  a  disabling  accident  to  the  insured 
(accident  insurance) ;  or  against  loss  to  property  insured  caused  by 
fire  (fire  insurance) ;  etc.  The  written  contract  between  the  insurer 
and  the  insured  is  called  a  policy.  The  premium  is  the  sum  paid 
to  the  insurer  for  the  risk  which  is  assumed,  and  is  estimated  at  a 
certain  %  of  the  amount  of  insurance,  though  frequently  expressed 
by  giving  the  cost  per  $100  or  per  $1000  of  the  risk  assumed. 
Thus,  f  %,  75  cents  per  $100,  or  $7.50  per  $1000  are  equivalent 
rates  expressed  in  different  forms. 

Identification  of  terms.  1.  The  sum  insured  is  the  base  or 
multiplicand.  2.  The  %  of  premium  is  the  rate  or  multiplier. 
3.  The  premium  is  the  percentage  or  product. 

NOTE  1.  The  product  of  the  base  and  the  rate  will  be  the  same,  irre- 
spective of  the  order  of  their  arrangement.  Hence,  for  convenience,  when 
the  rate  of  insurance  is  expressed  at  so  much  per  $100  or  per  $1000,  the  usual 
position  of  the  base  and  rate  is  reversed,  and  the  rate  per  $100  or  per  $1000 
is  taken  as  the  multiplicand,  and  the  number  of  hundreds  or  of  thousands  of 
dollars  is  taken  as  the  multiplier.  Thus,  the  premium  for  $8000  of  risk  at 
$6.25  per  $1000,  would  be  computed  at  8  times  $6.25,  or  $50. 

NOTE  2.  Fire  and  marine  policies  contain  an  "average  clause"  under 
which,  if  the  property  is  only  partially  insured  and  is  totally  destroyed  within 
the  period  of  insurance,  the  insurer  is  held  responsible  for  the  total  sum  in- 


224  PERCENTAGE 

sured;  but  if  the  property  is  only  partially  destroyed,  the  insurer  is  only 
liable  for  such  a  part  of  the  loss,  as  the  sum  for  which  the  property  is  insured 
is  of  the  full  value  of  the  property. 

EXAMPLES 

1.  A  man  insured  his  house  for  $6800,  the  rate  of  insurance 
being  f  %.     What  premium  did  he  annually  pay? 

Sum  insured  or  base  (6800)  X  rate  of  prem.  (f  %)  =  prem.  ($42.50). 

2.  A  house  is  insured  at  J  %.     What  was  the  sum  insured  if 
the  premium  paid  is  $26.25? 

3.  A  house  is  insured  for  $8200  and  the  premium  paid  is 
$71.75.     What  is  the  rate  of  insurance? 

4.  A  house  is  insured  for  $7500.     If  the  rate  of  insurance  is 
|  %,  what  premium  did  he  annually  pay? 

5.  A  ship  was  insured  for  $75300,  the  rate  of  insurance  being 
f  %.     What  was  the  premium  paid? 

6.  A  man  had  his  life  insured  for  $9000  at  the  annual  rate  of 
$35.50  per  $1000.     What  annual  premium  does  he  pay?  [Note  1] 

7.  A  building  valued  at  $12000  and  insured  for  $8000  was 
damaged  by  fire.     If  the  appraisement  of  the  damage  was  $2400, 
what  sum  was  received  from  the  insurer  in  settlement?  [Note  2] 

8.  A  stock  of  merchandise  valued  at  $40000  was  insured  in 
one  company  for  $20000  and  in  a  second  company  for  $10000. 
If  the  stock  was  damaged  by  fire  and  water  at  an  appraisement 
of  $6000,  what  proportion  of  the  loss  should  be  paid  by  each 
company? 

9.  A  steamboat  valued  at  $120000  was  insured  in  one  com- 
pany for  $20000,  in  a  second  company  for  $30000,  and  in  a  third 
company  for  $40000.     If  a  loss  of  $15000  was  sustained,  how  much 
insurance  should  the  owners  receive  from  each  company. 

TAXES 

287.  A  tax  is  an  assessment  levied  for  public  purposes  upon 
the  person  (poll  tax  or  capitation  tax),  upon  the  income  (income 
tax),  or  upon  the  property  (property  tax),  of  the  inhabitants  of 
a  country. 

Identification  of  terms.     1.  The  assessed  value  of  the  property, 


DUTIES   OR   CUSTOMS  225 

or  the  amount  of  income  when  over  a  prescribed  minimum,  is  the 
base  or  multiplicand.  2.  The  rate  of  taxation  is  the  rate  or  mul- 
tiplier. 3.  The  tax  is  a  percentage  or  product. 

NOTE.  If  the  rate  of  taxation  is  expressed  at  so  much  on  $100,  the  com- 
putation may  be  made  as  explained  in  Note  1,  286;  or  it  may  be  reduced  to 
an  equivalent  rate  %  (cents  on  $1)  by  moving  the  decimal  point  of  the  rate 
on  $100  two  places  to  the  left.  Thus,  $1.45  on  $100  =  $.0145  on  $1,  or  1.45  %. 

EXAMPLES 

What  is  the  tax,  if  the  assessed  value  of  the  property  is 

1.  $8500  and  the  rate  of  taxation  $1.32  on  $100? 

$1.32  on  $100  =  1.32  %  [Note].  Hence,  assessed  value  or  multiplicand 
($8500)  X  specific  rate  of  taxation  or  multiplier  (.0132)  =  percentage  of  tax 
or  product  ($112.20). 

2.  $16200  and  the  rate  of  taxation  £  %? 

3.  $9500  and  the  rate  of  taxation  $1.65  on  $100? 

4.  $12300  and  the  rate  15  mills  on  $1? 
6.  $8700  and  the  rate  of  taxation  1J%? 

6.  $23600  and  the  rate  $1.72  on  $100? 

7.  $17200  and  the  rate  8  mills  on  the  dollar? 

8.  My  real  and  personal  property  are  assessed  at  $5300. 
What  tax  will  I  be  required  to  pay  if  the  rate  of  property  tax 
is  75  cents  on  $100,  and  my  poll  tax  is  $1? 

9.  If  the  assessed  value  of  the  taxable  property  of  a  county 
is  $13578200  and  the  sum  required  by  taxation  is  $204538,  what 
should  be  the  rate  of  taxation  on  $100,  if  there  are  1730  resi- 
dents who  are  subject  to  a  poll-tax  of  50  cents? 

10.  If  the  assessed  value  of  real  estate  in  a  town  is  $758200, 
and  of  personal  property  $215000,  and  the  sum  required  by  taxa- 
tion is  $14469,  and  4%  of  the  tax  upon  personal  property  is 
estimated  to  be  uncollectible,  what  should  be  the  rate  of  taxation 
expressed  in  mills  on  $1? 

DUTIES  OR  CUSTOMS 

288.  INTRODUCTION.  Duties  or  customs  are  taxes  levied  by 
a  nation  upon  goods  imported  from  foreign  countries.  By  the 
Payne-Aldrich  Tariff  Act  of  1909,  these  duties  on  some  articles 
are  fixed  at  a  certain  per  cent,  of  their  market  value  at  the  place 


226  PERCENTAGE 

of  purchase,  including  the  value  of  the  container  or  covering  in 
which  the  goods  are  shipped  unless  such  container  or  covering  is 
otherwise  specially  subject  to  duty  under  some  other  provision 
of  that  tariff  act;  and  on  other  articles  are  fixed  upon  the  weight 
or  measure  of  imported  goods  without  reference  to  their  value. 
When  levied  at  a  certain  per  cent,  of  their  value,  duties  are  called 
ad  valorem;  and  when  levied  at  so  much  per  yard,  per  pound,  per 
gallon,  etc.,  they  are  called  specific.  Sometimes,  upon  the  same 
article,  both  an  ad  valorem  and  a  specific  duty  may  be  levied;  as 
jute  matting  15  cents  per  square  yard  and  30%  ad  valorem. 

The  ports  at  which  duties  are  collected  are  called  ports  of  entry; 
the  public  building  in  which  business  relating  to  customs  is  trans- 
acted is  termed  a  custom  house;  and  the  chief  official  in  charge  of 
the  customs  is  named  the  collector  of  customs. 

IDENTIFICATION  OF  TERMS 

Ad  valorem  duties.  1.  The  market  value  at  the  place  of  pur- 
chase, reduced  to  United  States  money,  is  the  base  or  multiplicand. 
2.  The  %  of  ad  valorem  duty  is  the  rate  or  multiplier.  3.  The  ad 
valorem  duty  is  the  percentage  or  product. 

Specific  duties.  1.  The  specific  duty  levied  on  1  unit  of  the 
quantity  subject  to  such  duty,  is  the  multiplicand.  2.  The  number 
of  units  subject  to  such  duty  is  the  multiplier.  3.  The  total  specific 
duty  is  the  product. 

NOTE  1.  In  reducing  foreign  monetary  units  to  United  States  money, 
use  Table  138;  and  when  estimating  the  ad  valorem  duty,  omit  any  fraction 
of  a  dollar  from  the  base  if  less  than  50  cents,  and  count  it  as  an  additional 
dollar  if  50  cents  or  more. 

NOTE  2.  In  reducing  tons  to  pounds,  or  the  reverse,  estimate  2240 
pounds  to  a  ton. 

EXAMPLES  FOR  PRACTISE 

Find  the  ad  valorem  duty  upon  foreign  goods  invoiced  at 

1.  £260  17s.  6d.,  allowing  5  %  for  leakage,  rate  of  duty  40  %. 
£260  17s.  6d.  =  £260.875  [160].     $4.8665  X  260.875  =  $1269.548  +  [138]. 

5  %  of  $1269.548  =  $63.477,  allowance  for  leakage.  $1269.548  -  $63.477  = 
$1206.071,  practically  $1206  [Note  1],  dutiable  value.  $1206  X  .40  =  $482.40, 
ad  valorem  duty. 

2.  12768.70  marks,  breakage  10  %,  duty  30  %. 

3.  9586.20  francs,  leakage  5  %,  duty  45  %. 

4.  £948  2s.  6d.,  duty  60  %. 


SIMPLE   PROPORTION  227 

5.  17294.40  milreis,  duty  25  %. 

6.  23742  yen,  duty  35  %. 

Find  the  specific  duty  upon  an  importation  of 

7.  213  tons  white  lead  at  2J  cents  per  pound. 

8.  345  gallons  cod-liver  oil  at  15  cents  per  gallon. 

9.  600  yds.  matting,  2  yds.  wide,  at  3J  cents  per  sq.  yd. 

10.  What  is  the  duty  upon  2400  yards  carpet  which  is  30 
inches  wide,  if  invoiced  at  12s.  6d.  per  yard,  and  subject  to  an 
ad  valorem  duty  of  40%,  and  to  a  specific  duty  of  28  cents  per 
square  yard? 

11.  What  is  the  duty  on  350  gross  lead  pencils  invoiced  at 
32  francs  per  gross,  if  the  ad  valorem  duty  is  25  %  and  the  specific 
duty  45  cents  per  gross? 

12.  What  is  the  duty  upon  an  importation  of  3700  yards  of 
27-inch  dress  goods  if  invoiced  at  7s.  6d.  per  yard,  and  subject 
to  an  ad  valorem  duty  of  50  %  and  a  specific  duty  of  8  cents  per 
square  yard? 

SIMPLE  PROPORTION 

289.  INTRODUCTION.  Proportion  is  the  name  of  that  process 
by  which  a  quantity  (the  required  answer)  is  found  which  will 
have  the  same  relation  to  a  similar  given  quantity  (expressing 
similar  units  to  the  required  answer),  as  the  given  quantity  of 
which  the  answer  is  required  has  to  the  similar  given  quantity 
of  which  the  same  specific  answer  is  known.  Thus,  if  40  bushels 
of  corn  cost  $30  and  it  is  required  to  find  the  proportionate  cost  of 
80  bushels  of  the  same  kind  of  corn,  the  required  cost  must  have 
the  same  relation  to  the  given  cost  ($30)  as  the  given  quantity  of 
which  it  is  required  to  find  the  cost  (80  bu.)  has  to  the  similar 
given  quantity  (40  bu.)  of  which  the  cost  is  known;  and  as  80 
bushels  of  which  the  cost  is  required  are  2  times  40  bushels  of 
which  the  cost  is  known,  so  the  required  cost  must  be  2  times 
the  known  cost  ($30),  or  $60.  The  preceding  process  is  appropri- 
ately called  a  proportion  because  the  incomplete  couplet  of  similar 
quantities  of  which  the  required  answer  is  one  term  ($30  to  ?)  is 
proportional  to  (have  the  same  relation  to  each  other  as)  another 
couplet  of  similar  quantities  of  which  both  terms  are  given  (40 
bu.  to  80  bu.). 

NOTE  1.  The  two  terms  of  a  proportion  which  are  of  the  same  name  are 
called  a  couplet  of  that  proportion,  and  every  problem  in  proportion  must  con- 


228  PROPORTION 

tain  at  least  two  couplets  (one  complete  couplet  and  one  incomplete  couplet). 
The  first  or  left-hand  term  of  each  couplet  is  distinguishably  called  its  ante- 
cedent (going  before);  and  the  second  or  right-hand  term,  its  consequent  (fol- 
lowing after).  The  relation  which  the  antecedent  of  any  couplet  has  to  its 
consequent  is  called  the  ratio  (the  relation)  of  that  couplet. 

NOTE  2.  A  problem  in  simple  proportion  is  one  in  which  a  complete 
couplet  of  similar  terms  is  given  which  have  a  certain  relation  to  each  other; 
and  in  which  one  term  of  a  second,  incomplete  couplet  of  other  similar  terms  is 
given  and  it  is  required  to  find  the  remaining  term  of  that  couplet  which  shall 
have  the  same  relation  to  its  given  term,  as  the  corresponding  terms  of  the 
complete  couplet  have  to  each  other. 

NOTE  3.  A  direct  proportion  is  one  in  which  the  required  answer  will  be 
proportionately  increased  as  the  term  of  which  the  answer  is  required,  is 
increased;  or  decreased  as  that  term  is  decreased.  Thus,  if  the  required 
result  is  wages  of  a  given  number  of  men,  the  greater  the  number  of  men 
employed,  the  more  the  wages,  and  the  fewer  the  number  of  men  employed, 
the  less  the  wages;  or  if  the  cost  of  a  given  number  of  articles  is  required, 
the  more  the  number  of  articles  bought,  the  greater  the  cost,  and  the  fewer 
the  number  of  articles,  the  less  the  cost,  etc. 

NOTE  4.  An  inverse  proportion  is  one  in  which  the  required  answer  will 
be  proportionately  diminished,  as  the  term  of  which  the  answer  is  required  is 
increased;  or  increased  as  the  term  of  which  the  answer  is  required,  is  de- 
creased. Thus,  if  the  required  answer  is  days  in  which  a  given  task  can  be 
performed  by  a  given  number  of  men,  it  is  evident  that  the  greater  the  number 
of  men  employed  upon  the  task,  the  fewer  will  be  the  number  of  days  to  finish 
it;  and  the  fewer  the  number  of  men  employed,  the  greater  the  number  of 
days  to  complete  it. 


ILLUSTRATIVE  EXAMPLES 

1.  If  56  pounds  of  butter  cost  $11.76,  what  will  32  pounds 
of  the  same  kind  of  butter  cost? 

SOLUTION  EXPLANATION.     The  required  cost  of  32  Ib.  has 

«      g         ^  the  same  relation  to  the  given  cost  of  56  Ib.  ($11.76) 

Gi  v  #9  —  <RA  79    as  tne  numDer  of  pounds  of  which  the  cost  is  required 
PXP/       RM^    (32)  hag  to  the  number  of  poimds  Of  which  the  cost 

fifi  is  given  (56).     As  32  Ib.  are  ff  of  56  Ib.,  so  the  re- 

H  quired  cost  of  32  Ib.  must  be  H  of  the  given  cost  of 

1  56  Ib.,  and  $11.76  X  !f  =  $6.72. 

2.  If  9  masons  can  build  a  wall  in  8  working  days,  how  many 
days  will  be  required  for  12  masons  to  build  the  same  wall? 


SIMPLE    PROPORTION  229 

SOLUTION  EXPLANATION.     More   masons    (than   9    masons) 

can  build  the  wall  in  proportionately  fewer  days;  that 

A     «  i          is,  as  the  number  of  masons  is  increased,  so  propor- 
?  =  i)aays  tionately  more  of  the  wall  will  be  built  in  l  day>  and 

1$  the  fewer  will  be  the  days  required  to  complete  the 

3  wall.     Hence,  12  masons  (*£•  or  f  of  9  masons)  can 

1  inversely  build  the  wall  in  the  inverse  of  |  of  8  days, 

that  is,  in  f  of  8  days,  or  6  days. 

RULE.  1.  Compare  the  given  term  of  which  the  answer  is  re- 
quired with  the  given  term  which  is  similar  to  it  in  name  and  con- 
cerning which  an  affirmation  is  made  which  is  similar  in  name  to 
the  required  answer;  and  express  the  result  of  the  comparison  in  the 
form  of  a  common  fraction,  inverting  its  terms  when  the  proportion 
is  seen  to  be  inverse  [Note  4].  2.  Multiply  the  given  term  which  is 
of  the  same  specific  name  as  the  required  answer  by  the  obtained 
fraction. 

EXAMPLES  FOR  PRACTISE 

1.  If  200  pounds  of  bacon  cost  $24,  what  will  475  pounds 
of  the  same  kind  of  bacon  cost? 

2.  If  575  yards  of  cloth  cost  $477.25,  what  will  340  yards 
of  the  same  cloth  cost? 

3.  If  $1125  are  required  weekly  to  pay  75  operatives  in  a 
factory,  how  much  will  be  required  weekly  if  40  additional  oper- 
atives are  employed  at  the  same  wages? 

4.  How  many  pounds  of  beef  will  feed  320  men  if  90  pounds 
are  sufficient  to  feed  80  men? 

5.  If  75  sheep  of  a  flock  were  sold  for  $292.50,  how  many 
sheep  from  the  same  flock  should  be  sold  at  the  same  rate  for 
$1053? 

6.  What  will  642  tons  of  coal  cost  if  49  tons  of  the  same  coal 
can  be  bought  for  $303.80? 

7.  If  52  pounds  of  ham  are  bought  for  $12.48,  what  should 
be  the  cost  of  75  Ib.  12  oz.  of  the  same  kind  of  ham? 

SUGGESTION.  If  one  term  of  a  problem  in  proportion  is  a  compound 
number,  it  should  first  be  reduced  to  a  simple  denominate  number  by  156  or 
160;  and  both  it  and  the  term  with  which  it  is  to  be  compared  should  express 
similar  denominate  units. 


230  PROPORTION 

8.  If  $520  will  produce  $10.40  interest,  how  much  interest 
will  $820  produce  if  loaned  for  the  same  time  and  at  the  same 
rate? 

SUGGESTION.  All  of  the  given  terms  of  a  proportion  may  be  of  the  same 
general  name  without  necessarily  being  of  the  same  specific  name.  Thus,  in 
Ex.  8,  $520  and  $820  are  specifically  dollars  of  principal;  and  $10.40  and  the 
required  answer  are  specifically  dollars  of  interest. 

9.  What  principal  will  produce  $17.82  interest  if  $3784  will 
produce  $104.06  interest  at  the  same  rate  per  annum  and  in  the 
same  time? 

10.  If  a  certain  principal  will  produce  $19.98  interest  in  296 
days,  in  what  time  will  the  same  principal  produce  $10.26  interest 
if  loaned  at  the  same  rate  per  annum? 

11.  A  house  30  feet  high  casts  a  shadow  18  feet  long.     How 
high  is  a  steeple  which  casts  a  shadow  75  feet  long  at  the  same 
time  in  the  day? 

12.  If  a  tree  50  feet  high  casts  a  shadow  of  10  feet,  how  long 
a  shadow  will  be  cast  at  the  same  moment  by  another  tree  which 
is  40  feet  high? 

13.  If  30  men  can  complete  a  given  task  in  40  days,  how  many 
men  are  necessary  to  complete  it  in  50  days? 

14.  In  how  many  days  can  75  men  build  a  bridge,  if  it  has 
been  estimated  that  125  men  can  build  it  in  18  days? 

15.  If  175  pounds  of  beef  are  sufficient  to  feed  a  gang  of 
laborers  for  5  days,  how  much  beef  will  be  necessary  to  feed  them 
for  15  days? 

16.  What  should  be  the  tax  on  a  house  assessed  at  $4500, 
if  a  house   in   the   same   town  which  is   assessed   at   $6000   is 
taxed  $75? 

17.  A  and  B  are  partners  in  business,  the  former  investing 
$5000  and  the  latter  $7000.     The  net  gains  during  a  certain  inter- 
val were  $3600,  which  were  divided  between  them  in  proportion 
to  their  respective  investments.     How  much  of  the  gain  should 
each  receive? 

18.  A  clothier  sold  a  suit  of  clothes  at  a  profit  of  $4.80  and 
gained  20%.    What  would  have  been  his   %  of  gain  if  he  had 
sold  the  suit  at  a  profit  of  $6? 


COMPOUND   PROPORTION  231 

19.  A  merchant  sold  goods  for  $644  and  gained  15  %.     What 
would  have  been  his  %  of  gain  if  he  had  sold  them  for  $756? 

20.  A  dealer  sold  merchandise  for  $45  and  gained  25%.     At 
what  price  must  he  have  sold  the  merchandise  to  gain  30  %? 


COMPOUND  PROPORTION 

290.  INTRODUCTION.  A  problem  in  compound  proportion  is 
one  in  which  an  answer  is  required  which  depends  upon  more  than 
one  couplet  of  similar  terms,  and  which  therefore  requires  as  many 
separate  comparisons  as  there  are  given  couplets  upon  which  the 
required  answer  depends.  Each  separate  comparison  is  made 
in  the  same  manner  as  in  simple  proportion;  and  the  resultant 
of  the  several  comparisons  (a  compound  fraction)  when  reduced 
to  a  simple  fraction,  will  express  what  proportion  of  the  given 
term  which  is  of  the  same  specific  name  as  the  required  answer, 
must  be  taken  to  obtain  the  required  answer. 

ILLUSTRATIVE  EXAMPLE 

If  72  men  can  dig  a  sewer  which  is  240  yards  long,  12  feet 
wide,  and  9  feet  deep,  in  18  days,  how  many  men  will  be  required 
to  dig  in  similar  soil  a  sewer  which  is  360  yards  long,  15  feet  wide, 
and  24  feet  deep,  in  12  days? 

SOLUTION  EXPLANATION.     If  72  men  are  required 

to  dig  a  sewer   240  yards  long,  the  other 
dimensions  being  equal,  it  will  require  ff § 

wa    t «     <M    t«     1  K  of  72  men  to  dig  a  sewer  36°  yards  lons 

3*  of  £§  of  &  Of  if  =  i£     [Note  3,  289].     If  72  men  are  required  to 

^      dig   a   sewer    12    ft.   wide,  the   remaining 

J2  dimensions  being  equal,  it  will  require  if 

of  72  men  to  dig  a  sewer  15  ft.  wide.      If  72 

men  are  required  to  dig  a  sewer  9  ft.  deep, 

=j  of  n  men  =  540  men       the  .oth£  dimensions  *><**  elual>  *  wil1 
^J  require  V  of  72  men  to  dig  a  sewer  24  ft. 

deep.      If  72  men  are  required  to  dig  a 

sewer  in  18  days,  the  dimensions  of  each  sewer   being   the  same,   it  will 
inversely  require  {I  of  72  men  to  dig  it  in  12  days  [Note  4,  289]. 

Each  of  the  above  separate  comparisons  shows  what  proportion  of  72  men 
would  be  required  if  there  were  no  other  element  entering  into  the  comparison; 
therefore  if  the  results  of  the  several  separate  comparisons  are  combined  into 


232  PROPORTION 

a  single  expression,  the  resultant  compound  fraction  (If  §  of  If  of  2g4  of  |l  of 
72  men),  or  simplified  by  reduction,  -^  of  72  men,  or  540  men  will  be  the  re- 
quired number  of  men  to  dig  a  sewer  360  yd.  long,  15  ft.  wide,  24  ft.  deep,  in 
12  da.,  if  all  the  elements  which  enter  into  the  comparison  are  jointly 
considered. 

RULE.  1.  Separately  compare  each  given  term  upon  which  the 
required  answer  depends  with  the  corresponding  given  term  of  the 
same  name,  of  which  an  affirmation  is  made  which  is  similar  in 
name  to  the  required  answer;  and  express  the  result  of  each  separate 
comparison  in  the  form  of  a  common  fraction. 

2.  Combine  each  separate  fraction  which  expresses  a  simple  pro- 
portion when  it  is  separately  considered  into  a  compound  fraction 
which  will  express  the  compound  proportion  when  all  the  terms  are 
jointly  considered;   and  reduce  the  resultant  compound  fraction  to  a 
simple  fraction  in  lowest  terms. 

3.  Multiply  the  given  term  which  is  of  the  same  specific  name 
as  the  required  answer  by  the  obtained  simple  fraction. 

EXAMPLES  FOR  PRACTISE 

1.  If  15  men  can  make  30000  shingles  in  5  days,  how  many 
shingles  can  25  men  make  in  9  days? 

2.  If  12  men  can  earn  $270  in  9  working  days,  how  much 
can  28  men  earn  in  5  working  days? 

3.  If  $864  loaned  for  126  days  at  5  %  per  annum  will  produce 
$15.12  interest,  what  principal  loaned  at  8%  for  54  days  will 
produce  $5.10  interest? 

4.  If  a  gain  of  $68  is  made  by  selling  merchandise  which  cost 
$340  at  20  %  profit,  what  will  be  the  proportionate  gain  from 
selling  merchandise  which  cost  $720  at  25  %  profit? 

5.  If  by  walking  4  miles  an  hour  for  9  hours  per  day  a  pedes- 
trian can  walk  a  certain  distance  in  10  days,  in  how  many  days 
can  he  walk  the  same  distance  at  the  rate  of  3  miles  an  hour  for 

8  hours  a  day? 

6.  If  a  contractor  charged  $336  for  digging  a  cellar  54  ft. 
long,  35  ft.  wide,  and  12  ft.  deep,  what  should  be  his  proportion- 
ate charge  for  digging  another  cellar  45  ft.  long,  30  ft.  wide,  and 

9  ft.  deep? 


EQUATION  OF  PAYMENTS 

291.  INTRODUCTION,    (a)  Equation,  as  a  general  term,  means 
the  averaging  or  equalizing  of  two  or  more  unequal  quantities 
[Ex.  13,  54].     (6)  When  applied  to  paj^ments,  it  is  the  process 
of  finding  one  date  at  which  a  single  payment  of  the  total  of  two 
or  more  debts  can  be  made  which  will  be  equal  to  paying  each  sep- 
arate debt  at  its  own  separate  due  date,     (c)  When  applied  to 
terms  of  credit  it  is  the  process  of  finding  one  uniform  term  of  credit 
for  all  the  items  of  a  purchase  or  a  sale  which  is  the  equivalent  of 
the  different  terms  of  credit  at  which  the  several  items  were  sep- 
arately purchased  or  sold. 

(d)  The  equivalent  date  at  which  two  or  more  payments,  due 
at  different  dates,  can  be  simultaneously  made  is  called  the  aver- 
age or  equated  date;  and  (e)  the  uniform  term  of  credit  which  is 
the  equivalent  of  two  or  more  different  terms  of  credit  is  called 
the  average  or  equated  term  of  credit. 

TO  FIND  THE  EQUATED  DATE  OF  PAYMENT  WHEN  THE 

DATES  OF  SALE  ARE  EQUAL  AND  THE  TERMS  OF 

CREDIT   UNEQUAL 

ILLUSTRATIVE  EXAMPLE 

292.  A  merchant  sold  goods   on  June  16,  1911,  as   follows: 
$600  at  90  days,  $800  at  30  days,  and  $700  at  60  days.     What 
is  the  equated  date  for  the  payment  of  the  sum  of  the  three  items? 

SOLUTION 

1911  The  use  of  The  use  of 

June  16         $600  for  90    da.   =    $54000  for  1  da. 
800   "   30     "     =      24000   " 
700   "   60     "     =      42000   "        " 
The  use  of  $2100  for  (?)  days  =  $120000   " 

$120000  -r-  $2100  =  57  times  1  da.,  or  57  days. 
June  16,  1911  +  57  da.  =  Aug.  12,  1911. 

EXPLANATION.  If  the  purchaser  had  been  required  to  pay  the  sum  ot 
the  three  items  ($2100)  on  the  day  of  purchase  (June  16),  he  would  have  been 
deprived,  as  per  terms  of  purchase,  of  the  use  of  $600  for  90  da.,  or  its  equiva- 
lent of  90  times  $600,  or  $54000  for  1  da.;  and  of  the  use  of  $800  for  30  da., 
or  its  equivalent  of  30  times  $800,  or  $24000  for  1  da. ;  and  of  the  use  of  $700 
for  60  da.,  or  its  equivalent  of  60  times  $700,  or  $42000  for  1  da.;  or  a  total 


234  EQUATION   OF   PAYMENTS 

deprivation  of  the  use  of  $54000  +  $24000  +  $42000,  or  $120000  for  1  da. 
Hence  the  equitable  time  for  the  payment  of  $600  +  $800  +  $700,  or  $2100, 
must  be  as  many  days  after  June  16  as  will  offset  this  deprivation.  As  the 
total  purchase  ($2100)  kept  57  days  will  as  nearly  as  possible  equal  this  de- 
privation (for  $2100  X  57  =  $119700,  which  is  as  near  a  product  to  $120000 
as  is  possible  with  an  integral  multiplier),  so  57  da.  after  June  16,  1911, 
or  Aug.  12,  1911,  must  as  nearly  as  possible  be  the  equated  date  of  payment. 

RULE.  Multiply  each  item  by  its  term  of  credit,  and  divide  the 
sum  of  the  resulting  products  by  the  sum  of  the  items,  to  find  the 
equated  term  of  credit. 

Add  the  equated  term  of  credit  to  the  date  of  purchase  to  find 
the  equated  due  date. 

NOTE  1.  As  the  day  is  the  lowest  unit  of  time  considered  in  terms  of 
credit,  if  the  division  is  not  exact,  disregard  the  remainder  if  less  than  half  the 
divisor,  or  regard  the  remainder  as  another  day  if  one-half  the  divisor  or  more. 

NOTE  2.  If  the  terms  of  credit  are  expressed  in  months,  and  the  division 
is  not  exact,  multiply  the  remainder  by  30  and  divide  the  resulting  product 
by  the  original  divisor,  to  reduce  the  remainder  to  days  or  SOths  of  a  month 
[86],  applying  Note  1  to  the  final  remainder. 

NOTE  3.  If  all  the  items  of  sale  are  already  equal,  they  may  be  dis- 
regarded; and  the  unequal  terms  of  sale  will  then  need  only  to  be  averaged 
[Ex.  13,  54];  but  if  the  items  are  also  unequal,  the  products  of  the  items  by 
their  respective  terms  of  credit  must  be  averaged. 

EXAMPLES  FOR  PRACTISE 
Find  the  equated  term  of  credit  of  the  following: 

(1)  (2)  (3) 

$500  at  3  months  $800  at  2  months  $525  at  3  months 

700   "  2       "  200  "  4       "  680  "  2 

400   "  4       "  900  "  1       "  720  "  4       " 

600  "  1       "  450  "  3       "  815  "  1       " 

Find  the  equated  date  of  the  following: 

(4)  (5) 

1912  1911 

Apr  16     $650  at  30  days  Oct.  23  $620  at  90  days 

"            470  "  90     "  "           400  "  30     " 

"            520  "  60     "  "           750  "  60     " 

"            350  "  90     "  "           800  "  90     " 


EQUATION   OF   PAYMENTS  235 

6.  A  house  was  sold  for   $6000,  one-third  payable  in  cash, 
one-third  in  6  months,  and  the  remainder  in  12  months,  without 
interest.     What  is  the  equated  time  for  paying  the  entire  pur- 
chase money?  [Note  3] 

7.  On  Mch.  23,  1911,  goods  were  sold  for  $645  at  60  days, 
for  $415  at  30  days,  and  for  $500  cash.     What  is  the  equated 
date  for  the  payment  of  the  entire  bill? 

8.  On  June  18,  1910,  goods  were  sold  to  a  customer  as  follows: 
$650  for  cash,  $800  at  60  days,  and  $400  for  cash.     At  what 
time  can  the  three  items  be  paid  without  loss  to  either  buyer 
or  seller? 


TO  FIND  THE  EQUATED  DATE  OF  PAYMENT  WHEN  THE 

DATES  OF  SALE  ARE  UNEQUAL  AND  THE  TERMS  OF 

SALE  ARE  EQUAL 

ILLUSTRATIVE   EXAMPLE 

293.  William  Garig  bought  of  George  W.  Keller  merchandise 
as  follows:  Aug.  24,  1912,  $862;  Sept.  16,  1912,  $547;  Oct.  28, 
1912,  $485.  If  no  credit  was  allowed,  what  was  the  equated 
date  for  paying  the  sum  of  the  three  items? 

FIRST   SOLUTION 

Assumed  date,  July  31,.  1912 

Days  from 

1912  assumed  Use  for  1  day  of 

date 

Auz  24  $862          X  24  -   \   3448  =  862  X  4  units 

'   1  1724    =  862  X  2  tens 

h   1ft     *A7          V  A7        i   3829  =  547  X  7  units 
Sept.  16    547          X47==  =  547  x  4  teng 


Oct  28      485          X  89  -       4365  =  485  X  9  units 

'   {3880    =  485  X  8  tens 

Use  of  $1894  for  (?)  da.  =    $89562  used  for  1  da. 
$89562  -v-  $1894  =  47  +  times  1  da.  after  assumed  date. 
Hence,  47  da.  after  July  31,  or  Sept.  16  =  equated  date. 

EXPLANATION.     For  convenience,  assume  as  the  required  equated  date, 
the  last  day  of  the  month  immediately  preceding  the  earliest  month  in  the 


236  EQUATION    OF    PAYMENTS 

account  (July  31);  note  the  extent  of  the  error  resulting  from  this  assumption, 
and  make  the  proper  correction. 

Thus,  if  the  first  item  ($862)  were  paid  on  this  assumed  date  (July  31), 
though  not  due  until  Aug.  24,  or  24  days  later,  the  purchaser  would  be  de- 
prived of  the  use  of  $862  for  24  da.,  or  the  equivalent  of  the  use  of  24  times 
$862  for  1  da. 

If  the  second  item  ($547)  were  paid  on  this  assumed  date  (July  31),  though 
not  due  until  Sept.  16,  or  47  da.  later,  the  purchaser  would  be  deprived  of  the 
use  of  $547  for  47  days,  or  of  the  equivalent  use  of  47  times  $547  for  1  da. 

If  the  third  item  ($485)  were  paid  on  this  assumed  date  (July  31),  though 
not  due  until  Oct.  28,  or  89  da.  later,  the  purchaser  would  be  deprived  of  the 
use  of  $485  for  89  da.,  or  of  the  equivalent  use  of  89  times  $485  for  1  da. 

Hence,  to  pay  the  sum  of  the  three  items  ($1894)  on  the  assumed  date 
(July  31)  would  cause  a  total  deprivation  to  the  purchaser  equivalent  to  the 
use  of  $89562  for  1  da. ;  and  to  prevent  this  inequitable  deprivation,  the  sum 
of  the  three  items  should  be  paid  as  many  days  after  July  31  as  will  offset 
this  loss,  that  is,  as  the  total  purchase  ($1894)  if  kept  47  da.  will  equal  the 
total  deprivation  ($89562  used  for  1  da.),  for  $1894  X  47  =  89018,  which  is 
as  near  that  deprivation  as  is  possible  to  obtain  with  an  integral  multiplier, 
so  47  da.  after  July  31,  or  Sept.  16,  must  as  nearly  as  possible  be  the  equated 
date. 

ABRIDGED   SOLUTION 

Assumed  date,  July  31,  1912 

Days  from        Hundreds 
1912  assumed          of  dollars 

date  for  1  da. 

A,^  o/L     «QRO  v  9/L  f    34  =  (8  hundred  X  4  units)+2  hundred  to  carry 

1 172  =  (86  tens  X  2  tens)  and  nothing  to  carry 
7         _  _       (    38  =  (5  hundred  X  7  units) +3  hundred  to  carry 

(  219  =  (54  tens  X  4  tens)  +  3  hundred  to  carry 
n  ,    Oft        xoc  v  QQ  f    44  =  (4  hundred  X  9  units) +8  hundred  to  carry 

(388  =  (48  tens  X  8  tens)  +  4  hundred  to  carry 
Use  of  $1894  for  (?)  da.  =  $89500  used  for  1  day. 
$89500  +  $1894  =  47  +  times  1  day  after  July  31,  or  Sept.  16. 

EXPLANATION.  Both  of  the  above  solutions  are  similar  in  principle;  but 
in  the  abridged  solution  much  unnecessary  labor  is  avoided  by  omitting  the 
multiplication  of  such  orders  in  the  items  as  will  not  affect  the  final  result; 
that  is,  by  actually  multiplying  only  such  orders  in  the  items  as  will  produce 
hundreds  of  dollars  for  1  day  (mentally  multiplying  one  or  two  of  the  next 
adjoining  rejected  orders  of  each  item  to  obtain  the  correct  carrying  figure  in 
hundreds,  counting  half  a  hundred  or  more  as  an  additional  hundred).  The 
use  of  (interest  of)  less  than  half  a  hundred  dollars  for  1  day  is  less  than  our 
lowest  unit  of  coinage  (Iff),  and  does  not  therefore  require  consideration. 


EQUATION    OF   PAYMENTS  237 

NOTE  1.  The  orders  which,  when  multiplied,  will  produce  results  in 
hundreds  of  dollars  for  1  day,  are 

Hundreds'  order  of  the  items  X  units'  order  of  the  days. 

Tens'  order  of  the  items  X  tens'  order  of  the  days. 

Units'  order  of  the  items  X  hundreds'  order  of  the  days. 

After  obtaining  the  sum  of  the  products,  annex  two  O's  to  make  that  sum 
express  its  true  value  in  even  hundreds  of  dollars,  and  then  proceed  as  in  the 
first  solution. 

RULE.  Assume  as  the  equated  date  the  last  day  of  the  month 
immediately  preceding  the  earliest  month  in  the  account.  Multiply 
each  item  by  the  interval  in  days  between  the  assumed  date  and  the 
date  of  such  item.  Divide  the  sum  of  the  obtained  products  by  the 
sum  of  the  items  to  find  the  number  of  days  to  count  forward  from 
the  assumed  equated  date  to  obtain  the  correct  equated  date. 

EXAMPLES   FOR  PRACTISE 
Find  the  equated  due  date  of  the  following  cash  sales: 

(1)  (2)  (3) 

1910,  Apr.  20  $800  1911,  Mch.  6  $416  1912,  July  3  $582 
May  13  750  May  19  300  Aug.  21  430 
June  28  675  June  6  52ti  Oct.  12  618 

(4)  (5)  (6) 

1911,  May  23  $275  1910,  Oct.  16  $275  1912,  Feb.  18  $468 
July  14  356  Nov.  28  350  Apr.  7  500 
Aug.  10  400  Dec.  2  196  May  23  260 
Sept.  21  540  Dec.  23  248  June  19  592 

NOTE  2.  If  all  the  items  have  a  uniform  term  of  credit,  and  it  is  expressed 
in  days,  first  find  by  the  preceding  process  the  equated  time  in  days  exclusive 
of  the  uniform  term  of  credit;  then  add  the  uniform  term  of  credit  to  the  result 
to  find  the  equated  time  inclusive  of  the  term  of  credit;  lastly  find  the  equated 
date  in  the  usual  manner. 

(7)  (8) 

1910,  May  23  $563  at  90  da.  1911,  July  14  $680  at  60  da. 
June  10  625   "          Aug.  25  560 
Aug.  2  718   "          Oct.  13  725 

NOTE  3.  If  the  uniform  term  of  credit  is  expressed  in  months,  first  find 
the  equated  date  exclusive  of  the  months  of  uniform  credit;  then  count  for- 
ward from  this  obtained  date,  as  many  calendar  months  as  are  contained 
in  the  uniform  term  of  credit. 


238  EQUATION    OF    PAYMENTS 

0)  (10) 

1912,  Jan.  17  $368  at  2  mo.  1910,  June  26  $680  at  3  mo. 
Mch.  8  415   "          Aug.  10  753 
May  23  298   "          Sept.  28  560 

(ID  (12) 

1911,  Apr.   24  $250  at  30  da.  1912,  Sept.  29  $175  at  1  mo. 
June  11     625        "  Oct.    18    256 

July   21     700        "  Nov.  10    340        " 

Aug.  14    290        "  Dec.     6    294 

TO  FIND  THE  EQUATED   DATE  OF  PAYMENT  WHEN  THE 

DATES  OF  SALE  AND  THE  TERMS  OF    CREDIT  ARE 

BOTH  UNEQUAL 

ILLUSTRATIVE  EXAMPLE 

294.  Find  the  equated  date  for  paying  the  total  of  the  fol- 
lowing sales:  1912  —  July  28,  $652.35  at  60  days;  Sept.  10, 
$471.20  at  30  days;  Oct.  13,  $537.63  at  90  days. 

FIRST   SOLUTION 

Assumed  date,  June  30,  1912 

1912  Use  for  1  day  of 


*    AA  J        «A*0  Q*  V      $  5          5219   =   <652  X  8  UnltS)  +  3 

July  28  at  60  da.  $652.35  X    8  J    5218g  =  (652  3  x  g 


471  90  V  1  09  942  =  (471  X  2  Unlts)  +  °  t0 

Sept.  10  at  30  da.   471.20 


(     2688  =  (537  X  5  units)  +  3  to  carry 
Oct.  13  at  90  da.     537.63  X  195  =          48387  =  (537.6  X  9  tens)  X  3  to  carry 

_  (    53763  =  (537.63  X  1  hundred) 

Use  of  $1661.18  for  (?)  da.  =  $210307  used  for  1  day. 
$210307  -T-  $1661  =  127  —  times  1  day  after  assumed  date. 
127  days  after  June  30,  1911  =  Nov.  4,  1912,  equated  date. 

EXPLANATION.  Assume  as  the  equated  date  the  last  day  of  the  month 
which  immediately  precedes  the  earliest  month  found  in  the  account  to  be 
averaged,  calculate  the  extent  of  the  error  if  the  assumption  proves  wrong, 
and  make  the  proper  correction,  as  follows: 

If  the  purchaser  had  paid  the  first  item  ($652.35)  on  June  30,  though  not 
due  until  60  da.  after  July  28,  or  88  da.  after  June  30,  he  would  have  been  de- 


EQUATION    OF   PAYMENTS  239 

prived,  as  per  terms  of  sale,  of  the  use  of  $652.35  for  88  da.,  which  would  have 
been  equivalent  to  an  error  of  the  interest  on  88  times  $652.35  for  1  da. 

If  the  second  item  ($471.20)  had  been  paid  on  June  30,  though  not  due 
until  30  da.  after  Sept.  10,  or  102  da.  after  June  30,  the  error  would  have  been 
equivalent  to  the  interest  on  102  times  $471.20  for  1  da. 

If  the  third  item  ($537.63)  had  been  paid  on  June  30,  though  not  due 
until  90  da.  after  Oct.  13,  or  195  da.  after  June  30,  the  error  would  have  been 
equal  to  the  interest  on  195  times  $537.63  for  1  da. 

Therefore,  if  the  sum  of  the  three  items  ($1661.18)  had  been  paid  on  the 
assumed  date  (June  30)  instead  of  on  their  respective  due  dates,  the  total 
error  would  have  been  the  sum  of  the  three  ascertained  errors,  or  the  interest 
on  $210307  for  1  da.;  and  to  correct  this  error  the  sum  of  the  three  items 
($1661.18)  should  be  kept  by  the  purchaser  as  many  days  after  June  30  as 
would  be  equivalent  to  the  use  of  $210307  for  1  da.  As  the  sum  of  the  products 
($210307)  divided  by  the  sum  of  the  items  ($1661.18)  equals  127  -,  so  $1661.18 
kept  127  da.  after  the  assumed  date  will  offset  this  loss;  and  127  days  after 
June  30,  or  Nov.  4,  1912,  must  be  the  required  date  upon  which  the  sum  of 
the  three  purchases  ($1661.18)  should  be  paid  without  involving  the  loss  of 
the  use  of  money  by  either  purchaser  or  seller. 

ABRIDGED   SOLUTION 

Assumed  date,  June  30,  1912 

Hundreds 


1912 

July  28  at  60  d,  S652.35  X    88 


Sept.  !0  at  30  da.    47L20X102      -  %  +  J 


(    27  (5  hund.  X  5)  +  2 
Oct.    13  at  90  da.    537.63  X  195      =  \  484  (53  tens  X  9)  +  7 

_  (  538  (537  units  X  1)  +  1 

Use  of  $1661.18  for  ?  da.  =  $210300  used  for  1  day. 
$210300  -f-  1661  =  127  times  1  da.  after  assumed  date. 
June  30,  1912  +  127  da.  =  Nov.  4,  1912,  equated  date. 

THE  EXPLANATION  of  the  abridged  solution  is  similar  to  that  of  the  first 
solution  preceding  it.  They  differ  only  in  the  method  of  performing  the 
multiplications.  The  contracted  method,  involving  no  products  of  less  value 
than  hundreds  of  dollars  for  1  day,  is  fully  explained  in  the  abridged  solution 
of  111.  Ex.,  293  and  Note  1  following  it. 


240  EQUATION    OF    PAYMENTS 

RULE.  Add  the  term  of  credit  of  each  item  to  the  multiplier  for 
that  item  as  obtained  by  Rule  293;  and  continue  in  accordance  with 
that  Rule. 

NOTE  1.  When  dividing  the  sum  of  the  products  by  the  sum  of  the  items, 
omit  the  cents  from  the  divisor  if  less  than  50,  or  increase  the  dollars  by  $1  if 
the  cents  are  50  or  more. 

EXAMPLES  FOR  PRACTISE 
Find  the  equated  date  of  payment  of  the  following: 

(1)  (2) 

1910,  Apr.  6  $600  at  90  da.  1911,  Mch.  14  $750  at  60  da. 
May  18  725  "  30  "        Apr.  25  825  "  90  " 
June  23  480  "  60  "         June  13  690  "  30  " 

(3)  (4) 

1912,  Feb.  17  $482.35  at  60  da.  1911,  Sept.  3  $248. 15  at  90  da. 

Apr.  29  527.60  "  30  "  Oct.  10  361.78  "  60  " 

June  2  368.45  "  90  "  Nov.  25  419.95  "  30  " 

July  13  416.25  "  30  "  Dec.  20  195.80  "  60  " 

(5)  (6) 

1909,  July  21  $682.40  at  30  da.  1912,  May  31  $425.89  at  60  da. 

Sept.  12  546.15  "  90  "  June  14  516.75  "  90  " 

Nov.  25  725.87  "  60  "  Aug.  3  398.78  "  60  " 

Dec.  4  498.78  "  90  "  Sept.  20  465.85  "  30  " 

NOTE  2.  If  a  term  of  credit  is  expressed  in  months,  it  has  been  found 
most  convenient  in  practise  to  proceed  as  follows:  —  If  goods  are  bought  on 
May  16  at  3  months,  it  is  readily  seen  that  3  months  after  May  16  will  include 
the  end  of  May  (31st),  the  end  of  June,  and  the  end  of  July  (31st),  or  2  Slsts 

2 

in  all;  hence  write  a  small  2  over  that  term  of  credit  (3  mo.),  and  when  it  is 
reached  in  the  equation,  it  will  be  identified  as  92  da.,  that  is,  as  3  months 
of  30  da.  each,  plus  the  increase  for  the  two  31sts.  Or,  if  an  item  at  2  months 
is  dated  Jan.  4,  1910,  the  term  of  credit  will  include  the  end  of  Jan.  (31st), 

the  end  of  Feb.  (28th),  a  net  minusage  of  1,  hence  write  —1  over  that  term 

-l 

of  credit  (2  mo.),  and  count  it  as  60  da.  less  1.  Or,  if  the  item  is  at  4  months, 
and  is  dated  Nov.  24,  1911,  the  term  of  credit  will  include  the  end  of  Nov., 

the  end  of  Dec.  (31st)  the  end  of  Jan.  (31st),  and  the  end  of  Feb.  (29th),  or  a 

l 
net  plusage  of  1,  expressed  4  months,  and  counted  120  da.  plus  1 ;  etc. 


AVERAGING  ACCOUNTS                                241 

(7)  (8) 

1912,  Apr.  16  $468.75  at  3  mo.  1911,  July  12  $728.15  at  2  mo. 

June  12     529.60  "  1    "  Sept.   3    465.70  "  4  " 

July  15    387.90  "  2   "  Oct.   18    598.67  "  1  " 

Aug.  28    625.75  "  4   "  Nov.  23    615.80  "  3  " 

(9)  (10) 

1911,  May  25  $568.35  at  3  mo.  1912,  Jan.   18  $396.75  at  2  mo. 

June    2    275.50  "  60  da.  Feb.  12    285.67  "  3  " 

Aug.  17     348.25  "  2  mo.  Apr.  18  2968.52  "  60  da. 

Sept.  13    465.72  "  30  da.  May  10      98.76  "  30  " 


AVERAGING  ACCOUNTS 

295.  INTRODUCTION.   The  process  of  averaging  an  account  is 
an  application  of  the  principles  of  equation  of  payments  to  an 
account  which  has  both  debit  and  credit  items.     The  method 
common  to  both  is  to  assume  a  date  of  payment,  to  ascertain  the 
resulting  error  by  comparison  of  the  due  date  of  each  item  with 
the  assumed  common  due  date  for  all  the  items,  and  to  make  the 
proper  correction.     When  all  the  items  were  debits,  it  was  seen 
that  the  assumption  of  a  date  of  payment  earlier  than  that  of  the 
several  items  would  uniformly  result  in  a  loss  to  the  debtor.     In 
averaging;  an  account,  however,  which  has  both  debit  and  credit 
entries,  the  resulting  error  from  assuming  a  date  of  settlement 
earlier  than  that  of  the  several  items  will  only  indicate  a  loss  to 
the  purchaser  with  respect  to  the  debit  entries;   and  will  show  a 
gain  to  him  with  regard  to  all  credit  entries.     The  net  gain  or  net 
loss  of  the  entire  account  must  therefore  be  the  determining  fac- 
tor in  finding  the  proper  correction  of  the  assumed  date. 

296.  Averaging  an    account  is  finding   the    proper   time   for 
paying  the  balance  of  an  account  which  includes  both  debit  and 
credit  entries. 

NOTE.  The  average  date  is  that  date  at  which  the  balance  of  an  account 
should  be  paid  to  avoid  the  loss  of  the  use  of  money  by  either  party  concerned 
in  the  account.  Hence,  if  the  balance  of  an  account  is  paid  before  the  average 
date,  it  is  usual  to  diminish  the  balance  by  the  compensatory  discount  thereon 
for  the  days  of  prepayment ;  and  if  paid  after  the  average  date,  to  increase  the 
balance  by  the  compensatory  interest  thereon  for  the  days  of  postpayment. 


242 


AVERAGING  ACCOUNTS 


TO  AVERAGE  AN  ACCOUNT  WITH  ITEMS  ON  BOTH  SIDES 

ILLUSTRATIVE  EXAMPLE 
Find  the  equated  date  for  paying  the  balance  of  the  following 


account  : 
Dr.                                                 J.  P.  SMITH                                                 Cr. 

1911 
May  23 

To  Mdse. 

$768 

42 

1911 
July  12 

By  Cash 

$694 

40 

July  16 

M 

537 

80 

Aug.  16 

M 

478 

75 

Aug.  27 

M 

659 

35 

Sept.  25 

M 

354 

80 

FIRST   SOLUTION 

Assumed  date,  April  30,  1911 


May  23  $768.42  X  23  = 


2305 
_ 15368 

Julyl6537.80X77=|3^ 


Aug.  27   659.35X119  = 


5934 


$137547 


$1965.57 
1527.95 
$437.62,  balance  of  items 


(     9083 

July  12  $694.40  X  73  = 

(  4ot)Uo 
(     ^8*30 

Aug.  16   478.75X108=    |4^ 

(  2838 

Sept.  25  354.80X148=  ]  14192 

(35480 


$1527.95  $154906 

137547 
bal.  of  products,      $17359 
$17359  -^  $438  =  40  -  times  1  da.  before  Apr.  30,  or  Mch.  21. 

EXPLANATION.  Assuming  April  30,  1911,  as  the  date  for  paying  all  the 
debit  items,  it  is  found  by  111.  Solution,  293,  that  if  Smith  had  paid  the  three 
debit  items  on  the  assumed  date  instead  of  their  respective  due  dates,  he 
would  have  lost  the  use  of  (the  interest  on)  $137547  for  1  da.;  and  considering 
the  credit  items  upon  the  same  assumption  that  the  debits  were  due  (and 
should  therefore  have  been  paid)  on  April  30,  it  is  found  that  Smith  would 
have  gained  the  use  of  $694.40  for  73  da.,  of  $478.75  for  108  da.,  and  of  $354.80 
for  148  da.,  or  a  total  gain  equivalent  to  the  use  of  $154906  for  1  da.  Hence, 
the  net  gain  to  Smith  upon  this  assumption  must  be  equivalent  to  the  use  of 


AVERAGING   ACCOUNTS  243 

$154906  (gain)  -  $137547  floss),  Or  to  the  net  use  of  $17359  for  1  da.  Smith 
should  therefore  pay  the  balance  of  his  account  ($437.62)  as  many  days  before 
the  assumed  date  (April  30)  as  will  offset  this  gain.  As  Smith's  net  gain 
($17359)  divided  by  Smith's  balance  ($437.62)  equals  40  -,  so  40  da.  before 
April  30,  or  March  21,  1911,  must  be  the  equated  date  upon  which  Smith 
should  pay  this  balance  without  loss  of  the  use  of  money  (of  interest)  by 
his  creditor  or  himself. 

ABRIDGED   SOLUTION 

Assumed  date,  April  30,  1911 

(    2*^  (    21 

May  23  $768.42  X    23  =  <  ^      July  12  $694.40  X    73  =  | 

(    38  ( 

July  16    537.80  X    77  =     3*°      Aug.  16   478.75  X  108  - 


Aug.  27   659.35  X  119  =  j   f     g        „    ,5.    Q      .,„_  ( 

(   I  Ad         Dtjpl/.    6U      Ocrx.OU   A.    IrtO  —  *\ 

$1965.57  $1375  ^355 

1527.95  $1527.95  $1549 

$437.62  1375 

$174 
$17400  +  $438  =  40  times  1  da.  before  Apr.  30,  or  Mch.  21. 

RULE.  Assume  as  the  equated  date,  the  last  day  of  the  month 
which  immediately  precedes  the  earliest  month  in  the  account.  Mul- 
tiply each  item  by  the  interval  in  days  between  the  assumed  date  and 
the  due  date  of  such  item;  and  divide  the  balance  of  the  resulting 
products  by  the  balance  of  the  items  to  find  the  error  in  the  assumed 
date.  Count  the  error  in  days  forward  from  the  assumed  date  if 
both  balances  are  on  the  same  side,  or  backward  from  the  assumed 
date  if  the  balances  are  on  opposite  sides. 

NOTE  1.  The  equated  date  for  paying  the  balance  of  an  account  does  not 
necessarily  mean  the  date  at  which  it  must  be  paid,  for  it  may  be  impossible 
for  the  debtor  at  that  date  to  know  that  it  would  be  so  payable,  as  in  the  pre- 
ceding 111.  Ex.,  in  which  the  equated  date  is  prior  to  the  date  of  any  of  the 
purchases,  perhaps  even  prior  to  the  debtor's  intention  to  make  any  purchase 
whatever.  In  other  instances,  the  debtor  may  have  the  requisite  knowledge 
of  the  equated  date  but  may  not  possess  the  means  on  the  equated  date  with 
which  to  pay  his  indebtedness.  In  either  case,  the  equated  date  is  the  proper 
time,  if  practicable,  for  paying  the  balance  of  the  account  without  loss  of  in- 


244 


AVERAGING   ACCOUNTS 


terest  to  either  debtor  or  creditor.  If  for  any  reason,  the  balance  is  paid 
before  the  equated  date,  the  debtor  is  usually  allowed  discount  on  said  balance 
for  the  days  of  prepayment,  and  if  paid  after  the  equated  date,  the  creditor 
will  usually  claim  interest  upon  said  balance  for  the  days  of  postpayment. 


Dr. 


(No.  1) 


EXAMPLES  FOR  PRACTISE 

WARREN  H.  SADLER 


Cr. 


1910 
Oct.   25 

To  Mdse. 

$675 

1910 
Nov.     2 

By  Cash 

$24900 

Dr.            (No.  2)                       GEO.  C.  ROUND                                             Cr. 

1911 

Mch.  18 

To  Mdse. 

$385 

90 

1911 
Apr.     8 

By  Cash 

$246 

75 

May  24 

14 

472 

75 

June  16 

u 

187 

35 

Dr.            (No.  3)                      E.  BENSEL  &  Co.                                            Cr. 

1910 
Apr.  17 

To  Mdse. 

$678 

35 

1910 
May    6 

By  Cash 

$263 

50 

June  12 

« 

594 

20 

July  18 

« 

345 

75 

July  24 

u 

725 

45 

Aug.  10 

« 

416 

80 

Dr.            (No.  4)                    HENRY  ALEXANDER                                           Cr. 

1912 
July     6 

To  Mdse. 

$416 

72 

1912 
Aug.    8 

By  Cash 

$327 

90 

Aug.  29 

tt 

532 

50 

Sept.  20 

ii 

615 

25 

Oct.    14 

it 

374 

65 

Nov.  12 

tt 

285 

45 

Dr.            (No.  6)                     H.  A.  GRIESEMER                                            Cr. 

1911 
May  13 

To  Mdse. 

$256 

15 

1911 
June   21 

By  Cash 

$243 

60 

July   16 

« 

476 

35 

Aug.    17 

a 

185 

75 

Aug.  30 

« 

348 

50 

Sept.  12 

« 

372 

50 

AVERAGING   ACCOUNTS                                245 

Dr.              (No.  6)                      A.  H.  ATWOOD                                              Cr. 

1910 
June     3 

ToMdse.,    60  da. 

$392 

70 

1910 
July    13 

By  Cash 

$243 

GO 

July  28 

30    " 

527 

98 

Aug.   25 

"   Note,  60  da. 

185 

75 

Sept.  19 

net 

476 

87 

Oct.    10 

"   Cash 

372 

50 

Dr.              (No.  7)                 FAIRMAN  A.  SADLER                                          Cr. 

1912 
Feb.  23 

To  Mdse.,    90  da. 

$485 

90 

1912 
Mch.    4 

By  Cash 

$275 

80 

Apr.  17 

60   " 

372 

65 

May   16 

"    Note,  60  da. 

300 

00 

May  26 

"            30    " 

427 

25 

July      8 

((           «        on     « 

250 

00 

Dr.              (No.  8)                  R.  M.  BROWNING                                          Cr. 

1911 
Oct.   24 

Dec.  18 
1912 
Jan.    20 

To  Mdse.,  net 
90  da. 
"          60    " 

$628 
473 
532 

50 

85 
05 

1911 
Nov.  12 
1912 
Jan.  21 

Feb.  18 

By  Note,  90  da. 

"        "      60    " 
"   Cash 

$352 
450 
275 

75 

00 
30 

Dr.              (No.  9)                        H.  C.  REITZ                                                Cr. 

1910 
Mch.  16 

To  Mdse.,  3  mo. 

$575 

80 

1910 
Apr.    28 

By  Note,  1  mo. 

$384 

45 

May  19 

net 

618 

35 

June     6 

"    Cash 

256 

85 

June  24 

2  mo. 

487 

15 

July    22 

"   Note,  3  mo. 

416 

75 

Dr.              (No.  10)                   W.  M.  CUTCHIN                                    .       Cr. 

1911 
Jan.    18 

To  Mdse.,  2  mo. 

$394 

67 

1911 
Feb.    23 

By  Note,  4  mo. 

$284 

62 

Feb.  20 

t(          ^    « 

472 

50 

Mch.  29 

«       (t     2    " 

316 

25 

Apr.     7 

"          3    " 

528 

75 

May   10 

"       "     3    " 

472 

38 

246                               AVERAGING   ACCOUNTS 
Dr.              (No.  11)                 C.  L.  REINDOLLAR                                           Cr. 

1912 
Feb.  29 

To  Mdse.,  60  da. 

$826 

45 

1912 
Mch.  27 

By  Cash 

$375 

90 

Mch.  18 

Net 

416 

20 

Apr.    12 

"   Note,  90  da. 

462 

75 

May  20 

"          3  mo. 

752 

68 

June  30 

"   Cash 

400 

00 

June  13 

90  da. 

528 

34 

July    18 

"   Note,  2  mo. 

275 

15 

12.  On  June  24,  1911,  H.  M.  Lee  bought  merchandise  for 
$753.25  and  90  days'  credit.     If  Lee  made  a  cash  payment  of 
$500  on  July  3,  1911,  what  is  the  equated  date  for  paying  the 
remainder? 

NOTE  2.  If  part  of  a  debt  is  paid  before  it  is  due,  many  merchants  permit 
the  debtor  to  postpone  payment  of  the  remainder  for  a  proportionate  time 
after  it  is  due.  The  date  of  payment  of  the  remainder  will  be  the  equated 
date,  found  in  the  usual  manner.  Such  an  extension,  however,  is  not  enf orcible 
by  law  if  the  creditor  objects. 

13.  On  July  18,  1912,  G.  M.  Hett  bought  merchandise  for 
$900  and  60  days'  credit;    and  on  Aug.  3,  1912,  paid  $300  on 
account.     How  much  did  he  owe  on  Dec.  13,  1912,  at  6%? 

NOTE  3.  It  is  customary  to  charge  interest  upon  the  balance  of  an  account 
for  the  number  of  days  it  is  paid  after  its  equated  date. 

14.  A  has  two  of  B's  notes;    one  of  $580,  dated  Mch.  24, 
1911,  payable  4  months  after  date;   and  the  other  of  $624,  dated 
Apr.  16,  1911,  payable  60  days  after  date.     On  May  6,  1911,  B 
paid  $700  to  A,  canceled  the  two  notes,  and  gave  a  new  30-day 
non-interest-bearing  note  for  the  remainder.     What  was  the  date 
of  the  new  note? 

NOTE  4.  The  note  must  be  so  dated  as  to  fall  due  on  the  equated  date,  at 
which  time  the  unpaid  balance  for  which  it  was  given  would  have  fallen  due. 

15.  Thomas  G.  Hayes  bought  goods  for  $650  on  Aug.  18, 
1911,  and  90  days'  credit.     On  Sept.  23,  1911,  he  paid  $175  on 
account;   and  on  Oct.  5,  1911,  he  paid  $225  and  gave  a  60-day 
interest-bearing  note  for  the  remainder.     What  should  have  been 
the  date  of  the  note? 

NOTE  5.  If  a  note  bearing  interest  from  its  date  is  given  in  settlement 
of  the  balance  of  an  account,  its  date  should  be  the  equated  date  (Note  3). 


AVERAGING  ACCOUNTS  SALES 


247 


AVERAGING  ACCOUNTS  SALES 

ILLUSTRATIVE  EXAMPLE 

297.  Find  the  equated  date  for  paying  the  net  proceeds  of 
the  following  sales  on  commission:  Sales,  1912  —  June  19,  $285.90; 
July  17,  $328.70;  Aug.  22,  $264.16.  Charges,  1912  — July  2, 
cash  advanced  consignor,  $350;  July  16,  freight,  $28.70;  July  30, 
drayage,  $8.25;  commission,  3%. 
SOLUTION 


Assumed  date,  May  31,  1912 


June  19  $285.90  X  19 


July  17    328.70  X  47 
Aug.  22    264.16  X  83 


26 
29 

J23 

131 


$878.76  $428 

$42800  +  $879  =  49  times  1  da. 
May  31  +  49  da.  =  July  19,  1912. 
$878.76  X  .03  =  $26.36  +,  com.  on  sales. 


EXPLANATION.  First  find  the 
equated  date  of  the  three  sales  by 
293,  obtaining  July  19,  1912,  as 
the  date  for  the  total  sales 
($878.76)  and  for  the  total  com- 
mission ($878.76  X  .03  =  $26.36). 

Next,  average  the  account  by 
296,  placing  all  the  charges  includ- 
ing the  commission  and  its  equated 
date,  on  the  debit  side;  and  plac- 
ing the  total  sales  ($878.76)  and 
its  equated  date  on  the  credit  side. 


Assumed  date,  June  30,  1912 


1912 
July  2 


$350.       X    2=  $7 
"  16  28.70  X  16  = 

"  30  8.25  X  30=    2 

"  19  (com.)  26.36  X  19= 
_ 

$413.31  19 


1912  ($79 

July  19  $878.76X19=) 

(       OO 

$167 

413.31  19 

Bal.  $465.45  $148  Bal. 

$14800  ^  $465  =  32  times  1  da. 
June  30  +  32  da.  =  Aug.  1,   1911, 


EXAMPLES  FOR  PRACTISE 

1.   An  agent  sold  a  consignment  as  follows:   1910  —  May  18, 
$7695.75;    June  2,  $5677.20;    June  25,  $4108.80.     His  charges 


248  CASH    BALANCE 

were:  1910  —  May  15,  freight,  $315;  May  20,  cash  advanced 
on  consignment,  $3500;  commission,  1J%.  Find  the  equated 
date  for  remitting  the  net  proceeds  to  his  principal. 

2.  Find  the  equated  date  for  paying  the  net  proceeds  of  the 
following  commission  sales:  Sales,  1911  —  Aug.  5,  $287.35;  Sept. 
2,  $328.49;   Oct.  20,  $289.30.     Charges,  1911  — July  28,  freight, 
$18.25;   Aug.  3,  cash,  $500;   commission,  2%. 

3.  Find  the  net  proceeds  of  the  following  commission  sales, 
and  the  equated  date  for  remitting  the  proceeds  to  the  consignor: 
Sales,    1912  — Apr.    17,    $583.75;    May   9,    $872.65;    June    12, 
$1075.80.     Charges,  1912  — Apr.  23,  drayage,  $12.50;    May  14, 
cash  advanced,  $1200;   commission,  2%. 

CASH  BALANCE 

298.  INTRODUCTION.  To  take  the  balance  of  any  account  is 
to  find  the  difference  between  its  total  debits  and  its  total  cred- 
its. To  prefix  "cash"  to  the  word  balance  adds  nothing  to  its 
significance;  for  all  balances  are  payable  in  cash  or  its  generally 
accepted  equivalent.  The  term  "cash  balance,"  in  business 
usage,  has  special  reference  to  the  difference,  on  the  date  of  tak- 
ing the  cash  balance,  or  date  of  adjustment,  between  the  present 
worth  of  all  the  debit  items  and  the  present  worth  of  all  the  credit 
items.  This  includes  consideration  of  the  debit  interest  on  each 
debit  item  that  falls  due  before  the  date  of  adjustment  for  the  num- 
ber of  days  between  its  due  date  at  which  it  is  legally  payable  to 
the  date  of  adjustment  when  it  is  assumed  to  be  paid;  and  the 
credit  interest  (discount)  on  each  debit  item  that  falls  due  after 
the  date  of  adjustment  for  the  number  of  days  between  the  date 
of  adjustment  at  which  it  is  assumed  as  being  payable,  to  its 
subsequent  due  date  at  which  it  is  legally  payable.  It  also  in- 
cludes consideration  of  the  credit  interest  on  each  credit  item,  if 
cash,  for  the  number  of  days  between  its  date  at  which  it  is  actu- 
ally paid  to  the  date  of  adjustment  at  which  it  is  assumed  to  be 
paid;  or  if  a  time  note  or  draft  maturing  before  the  date  of  adjust- 
ment, of  credit  interest  for  the  number  of  days  between  its  legal 
due  date  at  which  it  is  actually  paid  to  the  date  of  adjustment  at 
which  it  is  assumed  to  be  paid;  or  if  a  time  note  or  draft  maturing 
after  the  date  of  adjustment,  of  debit  interest  (discount)  for  the 
number  of  days  between  the  date  of  adjustment  at  which  it  is 
assumed  to  be  paid  and  its  legal  due  date  at  which  it  is  actually 
paid. 


CASH   BALANCE 


249 


299.  The  cash  balance  of  an  account  is  the  difference,  on  the 
day  of  adjustment,  between  the  present  worth  of  all  its  debit 
items  and  the  present  worth  of  all  its  credit  items. 

ILLUSTRATIVE  EXAMPLE 

Find  the  cash  balance  of  the  following  account  on  Dec.  13, 
1911,  at  6%. 

Dr.  JOSHUA  LEVERING  in  account  with  J.  H.  TYLER  Cr. 


1911 

Days 

Int. 

Amount 

1911 

Days 

Int. 

Amount 

July   28 

To  Mdse.,  3  mo. 

$268 

75 

July  20 

By  Note,  2  mo. 

$175 

83 

Aug.   12 

30  da. 

327 

13 

Sept.    2 

"    60  da. 

207 

27 

Oct.    19 

Net 

295 

84 

Oct.    28 

"   Cash 

148 

75 

Dr. 


SOLUTION 


Cr. 


1911 

Days 

Int. 

Am'nt 

1911 

Days 

Int. 

Ainou 

July    28 

To  Mdse.,  3  mo. 

46 

$2 

06 

$268 

7f> 

July   20 

By  Note,  2  mo. 

84 

$2 

40 

$17f 

Aug.   12 

30  da. 

93 

5 

07 

327 

13 

Sept.    2 

"       "     60  da. 

42 

1 

15 

207 

Oct.    19 

Net 

55 

2 

71 

295 

84 

Oct.    28 

"   Cash 

46 

1 

14 

148 

Dec.   13 

Bal.  of  Int. 

4 

79 

Dec.   13 

"  Bal.  of  Int. 

4 

79 

'  % 

364 

$9 

84 

$890 

•>1 

$Q 

84 

$896 

EXPLANATION.  Write  in  the  days'  column  opposite  each  item,  the  num- 
ber of  days  intervening  between  the  due  date  of  that  item  and  the  date  of 
taking  the  cash  balance  (Dec.  13).  Next  write  in  the  interest  column,  the 
interest  on  each  item  for  the  days  specified  in  the  days'  column. 

Then  write  the  difference  between  the  total  debit  interest  and  the  total 
credit  interest  ($4.79)  on  the  less  side,  thus  balancing  the  interest  columns; 
and  as  this  interest  balance  shows  that  Levering  owes  Tyler  $4.79  more 
interest  than  Tyler  owes  Levering,  enter  it  as  an  additional  item  in  the  amount 
column  on  the  debit  side  of  Levering's  account. 

Lastly,  write  the  difference  between  the  total  of  the  amount  column  on 


250 


CASH   BALANCE 


the  debit  side  and  the  total  of  the  amount  column  on  the  credit  side  ($364.66), 
on  the  less  side,  as  the  required  "cash  balance"  on  Dec.  13,  1911. 

RULE.  1.  Find  the  interest  of  each  item  of  the  account  from  its 
due  date  to  the  date  of  taking  the  cash  balance.  2.  Balance  the 
interest  columns.  3.  Carry  the  interest  balance  to  the  main  money 
column  on  the  side  opposite  to  that  in  which  it  is  written  in  the  interest 
column.  4.  Balance  the  main  money  columns. 

NOTE  1.  The  interest  of  items  which  fall  due  after  the  date  of  adjusting 
the  cash  balance  should  be  computed  for  the  number  of  days  between  the 
date  of  adjustment  and  the  due  date  of  such  items;  and  this  interest  should 
be  written  with  red  ink  in  the  interest  column  on  the  same  side  as  the  item; 
but  when  the  totals  are  taken,  this  red  interest  should  be  excluded  from  the 
total  interest  of  the  side  on  which  it  is  written,  and  included  with  the  total 
interest  of  the  opposite  side  where  it  properly  belongs. 

NOTE  2.  The  face  of  an  interest-bearing  note  should  be  regarded  as  the 
equivalent  of  so  much  cash  paid  on  the  day  it  commenced  to  draw  interest; 
and  the  face  of  a  non-interest-bearing  note  should  be  treated  as  the  equivalent 
of  so  much  cash  paid  on  its  due  date. 


EXAMPLES  FOR  PRACTISE 

1.   Find  the  cash  balance  of  the  following  on  Dec.  16,  1911, 

at  6  %. 

Dr.  EDMUND  BERKELEY  in  account  with  J.  B.  T.  THORNTON  Cr. 


1911 

Days 

Int. 

Amount 

1911 

Days 

Int. 

Amou 

July   18 

To  Balance 

*** 

** 

** 

$416 

70 

Aug.  20 

By  Cash 

*** 

* 

** 

$416 

Sept.  24 

"   Mdse. 

** 

* 

** 

286 

45 

Sept.  28 

„       „ 

** 

* 

** 

157 

Oct.      2 

,, 

** 

* 

** 

341 

60 

Oct.    18 

„       « 

** 

* 

** 

275 

Nov.  20 

„       „ 

** 

* 

** 

298 

40 

Dec.     1 

„       ,, 

** 

** 

198 

Dec.   16 

"  Bal.  of  Int. 

* 

** 

Dec.  16 

"   Bal.  of  Int. 

* 

** 

"      "     "  % 

** 

** 

**** 

** 

** 

** 

**** 

2.   Find  the  cash  balance  of  the  following  on  Oct.  16,  1912, 
at  5  %. 


CASH   BALANCE                                       251 

Dr.                                 JOB.  J.  JANNEY  in  account  with  N.  BRAWNER 

Cr. 

1912 

Days 

Int. 

Amount 

1912 

Days 

Int. 

Amount 

Mch.  16 

To  Mdse.,  3  mo. 

*** 

* 

** 

$298 

75 

Apr.     7 

By  Note,  90  da. 

** 

* 

** 

$178 

25 

May  28 

"       60  da. 

** 

• 

** 

328 

76 

July     3 

"       "       2  mo. 

** 

* 

** 

210 

38 

July    16 

"       Net 

** 

* 

** 

265 

83 

Aug.     8 

"   Cash 

** 

* 

** 

158 

16 

Oct.    16 

"   Bal.  of  Int. 

* 

** 

Oct.    16 

"   Bal.  of  Int. 

* 

** 

*** 

** 

Oct.    16 

% 

*** 

** 

** 

** 

*** 

** 

** 

** 

*** 

** 

3. 

Find  the  cash  balance  on  Nov.  18,  1911,  at  6%. 

Dr.                                      T.  C.  WILL  in  account  with  W.  P.  JORDAN                                    Cr. 

1911 

Days 

Int. 

Amount 

1911 

Days 

Int. 

Amount 

Mch.  21 

To  Mdse.,  30  da. 

*** 

** 

** 

$385 

112 

Apr.     7 

By  Note,  30  da. 

*** 

* 

** 

$193 

75 

May  10 

"        "       3  mo. 

*** 

* 

** 

278 

50 

May  20 

"       "     3  mo. 

** 

* 

** 

$216 

•25 

July   28 

"       Net 

*** 

* 

** 

328 

65 

July     7 

"   Cash 

*** 

* 

** 

265 

18 

Nov.  18 

"   Bal.  of  Int. 

* 

** 

Nov.  18 

"   Bal.  of  Int. 

* 

** 

Nov.  18 

% 

*** 

** 

** 

** 

**** 

** 

** 

** 

**** 

** 

4. 

Find  the  cash  balance  on  Nov.  8,  1912,  at  6  %. 

Dr.                                    WALTER  WATT  in  account  with  NEAL  FAHR                                     Cr. 

1912 

Days 

Int. 

Amount 

1912 

Days 

Int. 

Amount 

July    13 

To  Mdse.,  Net 

*** 

* 

** 

$285 

96 

Aug.  20 

By  Cash 

** 

* 

** 

$150 

35 

Sept.  23 

"       60  da. 

** 

** 

346 

75 

Sept.  16 

"   Note,  30  da. 

** 

** 

210 

50 

Oct.      3 

"        "       3  mo. 

** 

* 

** 

417 

30 

Oct.    18 

4  mo. 

*** 

* 

** 

168 

73 

Nov.    8 

"   Bal.  of  Int. 

* 

** 

Nov.    8 

"    Bal.  of  Int. 

* 

** 

Nov.    8 

% 

*** 

** 

* 

" 

**** 

** 

* 

**! 

**** 

4-* 

252  PARTNERSHIP 

5.  Arrange  the  following  items  in  the  form  of  an  account 
current,  ruled  as  in  the  preceding  examples,  and  find  the  cash 
balance  on  Sept.  5,  1912,  at  5%:  John  King  bought  of  J.  Malone 
Gaulden,  as  follows:  May  17,  1912,  mdse.  at  30  da.,  $378.25; 
June  28,  1912,  mdse.  at  4  mo.,  $296.52;  July  10,  1912,  mdse. 
net,  $419.75;  and  King  was  credited  on  June  28,  1912,  for  his 
note  of  $198.63,  at  90  da.;  on  July  5,  1912,  for  his  note  of  $128.46 
at  2  mo.,  and  on  July  20,  1912,  for  a  cash  payment  of  $293.75. 


PARTNERSHIP 

300.  INTRODUCTION,  (a)  A  business  partnership  is  a  com- 
bination of  the  resources  of  two  or  more  persons  for  the  estab- 
lishment or  the  maintenance  of  a  business  enterprise.  The  amount 
of  money,  or  property,  or  time,  or  the  character  of  the  duties  to 
be  assumed  by  each  partner,  or  the  manner  of  distributing  the 
gains  or  losses,  or  any  other  necessary  detail,  is  stipulated  in  an 
instrument  of  writing  called  a  partnership  agreement.  Such  a 
combination  of  persons  is  known  as  a  firm,  a  house,  or  a  company; 
and  each  person  in  the  firm  is  called  a  partner,  (b)  The  aggre- 
gate of  what  each  partner  contributes  to  the  partnership  is  termed 
his  investment;  and  the  difference  between  his  total  investments 
and  his  total  withdrawals  is  called  his  net  investment,  (c)  The 
sum  of  what  all  the  partners  contribute  is  known  as  the  capital 
of  the  firm;  (d)  and  the  difference  between  what  all  the  part- 
ners have  invested  and  what  all  the  partners  have  withdrawn 
plus  the  net  gain  or  minus  the  net  loss  since  the  last  partnership 
adjustment,  is  called  the  firm's  net  capital. 

(e)  The  assets  or  resources  of  a  firm  are  its  entire  possessions, 
including  obligations  due  to  it  from  others;  and  (/)  its  liabilities 
are  the  obligations  due  by  the  firm  to  others,  (g)  If  a  firm's  re- 
sources exceed  its  liabilities,  it  is  said  to  be  solvent;  and  the  excess 
is  known  as  the  firm's  net  solvency  or  net  capital;  but  (h)  if  a  firm's 
liabilities  exceed  its  resources,  it  is  said  to  be  insolvent,  and  the 
excess  is  known  as  the  firm's  net  insolvency  at  that  time. 

(i)  Due  allowance  being  made  for  possible  investments  or 
withdrawals,  if  any,  within  the  settlement-period,  the  net  gain  of 
a  firm  is  the  excess  of  its  net  capital  at  the  end  of  that  period  over 
its  net  capital  at  the  beginning  of  that  period;  (j)  or  the  decrease 
of  its  net  insolvency  at  the  end  of  that  period  below  its  net  insol- 
vency at  the  beginning;  (k)  or  the  sum  of  its  net  capital  at  the 
end  of  that  period  and  its  net  insolvency  at  the  beginning. 


PARTNERSHIP  253 

(Z)  Due  allowance  being  made  for  investments  or  withdrawals, 
if  any,  within  the  settlement-period,  the  net  loss  of  a  firm  is  the 
deficiency  of  its  net  capital  at  the  end  of  that  period  below  its  net 
capital  at  the  beginning  of  that  period;  (ra)  or  the  excess  of  its 
net  insolvency  at  the  end  of  that  period  above  its  net  insolvency 
at  the  beginning;  (n)  or  the  sum  of  its  net  insolvency  at  the 
end  of  that  period  and  its  net  capital  at  the  beginning. 

(o)  It  is  thus  seen  that  partnership  adjustments  involve  con- 
sideration of  three  terms:  first,  the  financial  condition  at  the 
beginning;  second,  the  financial  condition  at  closing;  third,  the 
net  gain  or  net  loss  during  the  interval,  (p)  These  three  terms 
have  a  fixed  relation  to  each  other,  additive  or  subtractive,  so 
that  if  any  two  of  them  are  given,  the  third  can  be  readily  found. 
(r)  Of  these  three  terms,  the  financial  condition  at  the  beginning 
of  the  settlement-period,  or  that  condition  as  modified  by  subse- 
quent investments  or  withdrawals,  if  any,  should  uniformly  be 
regarded  as  the  natural  standard  of  comparison,  as  follows: 

301.  To  find  the  gain  or  loss  of  a  firm,  (a)  Compare  its 
financial  condition  (solvency  or  insolvency)  at  the  end  of  a 
settlement-period  with  its  financial  condition  at  the  beginning  of 
that  period,  or  with  that  condition  as  modified  by  subsequent 
investments  or  withdrawals,  if  any,  within  that  settlement-period; 
and  (b)  if  its  financial  condition  at  the  end  of  the  period  is  found 
to  be  better  than  its  condition  at  the  beginning  of  the  period,  there 
has  been  a  gain;  if  worse,  a  loss;  and  the  amount  of  gain  or  loss 
will  be  the  difference  between  these  two  conditions. 

NOTE  1.  As  the  gain  or  loss  is  the  difference  between  two  financial  con- 
ditions, and  any  difference  added  to  the  less  term  will  equal  the  greater,  so, 
if  a  gain  or  loss  is  given,  and  the  financial  condition  at  the  close  of  the  settle- 
ment-period is  required,  it  will  be  the  gain  better,  or  the  loss  worse  than  the 
net  financial  condition  at  the  beginning;  or  if  the  net  financial  condition  at 
the  beginning  is  required,  it  will  be  the  gain  worse,  or  the  loss  better  than 
that  at  the  end. 

NOTE  2.  The  difference  between  similar  conditions,  that  is,  between  one 
net  capital  and  another  net  capital,  or  between  one  net  insolvency  and  another 
net  insolvency,  is  found  by  subtracting  the  less  from  the  greater;  but  the  dif- 
ference between  opposite  conditions,  that  is,  between  a  net  capital  and  a  net 
insolvency,  or  vice  versa,  is  found  by  addition. 

NOTE  3.  A  capital  at  the  end  of  a  period  is  better  than  at  the  beginning 
when  it  is  greater,  and  it  is  worse  when  it  is  less;  while  an  insolvency  at  the 
end  of  a  period  is  better  than  at  the  beginning,  when  it  is  less,  and  it  is  worse 
when  it  is  greater. 


254  PARTNERSHIP 

NOTE  4.  A  financial  condition  may  be  given  directly  as  so  much  capital 
or  insolvency;  or  it  may  be  given  indirectly  by  enumerating  the  different 
resources  and  liabilities,  or  by  access  to  the  books  of  account  from  which  the 
several  resources  and  liabilities  may  be  obtained. 

EXAMPLES  FOR  PRACTISE 
Find  the  gain  or  loss  : 

1.  Capital  at  beginning  $9285;   capital  at  closing  $12345. 

2.  Capital    at    beginning    $15275.80;     capital    at    closing 
$12685.92. 

3.  Insolvency  at  beginning  $9685.75;    insolvency  at  closing 
$13472.65. 

4.  Insolvency  at  beginning  $18271.36;  insolvency  at  closing 
$14275.36. 

6.   Capital    at   beginning  $7286.50;     insolvency   at  closing 
$2615.25. 

6.  Insolvency  at  beginning   $4586.25;     capital   at    closing 
$6718.60. 

Find  the  capital  or  insolvency  at  closing : 

7.  Capital   at   beginning   $12856.25;    gain   during  interval 
$4687.50. 

8.  Capital    at    beginning    $7268.42;    loss    during    interval 
$2178.65. 

9.  Insolvency  at  beginning  $5829.65;    gain  during  interval 
$6783.75. 

10.  Insolvency  at  beginning  $9685.30;    gain  during  interval 
$2785.90. 

11.  Capital    at    beginning    $3568.20;     loss    during    interval 
$6580.65. 

12.  Insolvency  at  beginning  $4689.75;    loss  during  interval 
$768.50. 

Find  the  capital  or  insolvency  at  beginning : 

13.  Capital  at  closing  $7856.35;  gain  during  interval  $2563.75. 

14.  Capital  at  closing  $3568.25;  gain  during  interval  $6528.15. 

15.  Capital  at  closing  $9275.80;  loss  during  interval  $1845.60. 

16.  Capital  at  closing  $1653.40;  loss  during  interval  $8215.25. 


PARTNERSHIP 


255 


at 
at 
at 
at 

closing 
closing 
closing 
closing 

$3720.65; 
$675.80; 
$1925.25; 
$2678.90; 

gain 
gain 
loss 
loss 

during 
during 
during 
during 

interval 
interval 
interval 
interval 

17.  Insolvency 
$1285.80. 

18.  Insolvency 
$4718.10. 

19.  Insolvency 
$658.25. 

20.  Insolvency 
$5815.60. 

21.  A  firm  commenced  business  with  the  following  resources: 
cash,  $2500;    merchandise,  $12500;    personal  accounts  due  the 
firm,  $1846.25;    bills  receivable,  $956.75.     Its  liabilities  at  the 
beginning  were:    due  to  others  for  merchandise  bought,  $3500; 
and  on  outstanding  notes,  $1256.18.     At  closing  they  had  re- 
sources as  follows :  merchandise  as  per  inventory  $15265.70;  cash, 
$780;   due  the  firm  on  personal  accounts,  $4625.80;   bills  receiv- 
able, $1475.50;  and  their  liabilities  were:  due  to  others  on  per- 
sonal accounts,   $1725.80;    and  on  outstanding  notes,   $865.70. 
What  was  the  net  gain  or  loss  of  the  firm  during  the  interval? 

22.  A  and  B  formed  a  copartnership,  the  former  investing  f 
of  the  capital  and  the  latter  J,  the  agreement  between  them  re- 
quiring the  gains  or  losses  to  be  divided  equally.     At  settlement 
the  net  gains  were  found  to  be  $6000,  their  resources  at  that  time 
being  merchandise  on  hand  $12500,  cash  $3500,  personal  accounts 
$4200,  bills  receivable  $1826;    and  their  liabilities,  outstanding 
notes  against  the  firm  $4500,   unsettled  bills  for  merchandise 
bought  $2700.     How  much  did  each  partner  invest  at  the  begin- 
ning? 

23.  A,  B,  and  C  engage  hi  business  as  partners,  with  a  capital 
of  $18000,  A  having  invested  A  of  the  capital,  B  T4T,  and  C  •&.     At 
closing,  the  firm's  books  showed  a  net  gain  of  $6420,  which  accord- 
ing to  agreement  was  to  be  shared  equally.     What  was  each  part- 
ner's interest  in  the  firm  at  settlement? 

DIVISION  OF  GAINS  OR  LOSSES  AMONG  PARTNERS  IN 
PROPORTION  TO  THEIR  ORIGINAL  INVESTMENTS 

302.  INTRODUCTION.  Unless  otherwise  specified  in  the  part- 
nership agreement,  the  courts  presume  that  the  gains  or  losses 
are,  by  implication,  to  be  divided  equally  among  all  the  partners, 


256  PARTNERSHIP 

irrespective  of  the  money  they  may  have  individually  invested, 
that  not  being  regarded  as  the  only  source  of  possible  gains.  What 
a  partner  may  lack  pecuniarily  may  be  offset  by  his  possession  of 
greater  skill  in  business,  greater  influence  over  patronage  in  the 
community,  or  by  assuming  greater  responsibility  of  management, 
etc.  In  the  examples  under  this  article,  it  is  assumed  that  the 
partners'  investments  were  made  at  commencing  business  and 
were  left  unchanged  by  subsequent  investments  or  withdrawals, 
and  that  inequalities  among  the  partners  in  other  respects  than 
money,  if  any,  have  been  provided  for  by  special  allowances 
(sometimes  improperly  called  salaries)  to  be  taken  from  the 
gross  gains  before  general  distribution  of  the  net  gain  is  made 
among  all  the  partners  in  proportion  to  their  respective  pecuniary 
investments  alone. 

ILLUSTRATIVE  EXAMPLE 

A  and  B  are  partners,  the  former  investing  $6000  and  the 
latter  $8000,  and  agreed  to  share  the  gains  or  losses  in  proportion 
to  their  respective  investments.  If  the  net  gain  during  the  first 
year  was  $5600,  what  was  the  share  of  each? 

SOLUTION  EXPLANATION.       Apply- 

$6000  +  S8000  =  S14000,  total  capital    **  * 


a.  a  a  a        af\<\(\  relation    to    the    total    gain 

-  <  '  ($5600)    as    A's 


($6000)  which  produced  his 
20  share    of    the    gain    has    to 

the  total  investment  ($6000 
8  400  +    $8000   =  $14000)  which 


X  produced     the     total     gain 

J4000  =  $3200>  B  s  snare  ($5600).     As  A's  investment 

($6000)  is  TYA%  or  f  of  the 


total  investment,  so  A's  share 

of  the  gain  should  be  f  of  the  total  gain  ($5600),  and  f  of  $5600  =  $2400. 
Proceed  in  a  similar  manner  to  find  B's  share  of  the  firm's  gain.     In  veri- 
fication, add  A's  share  of  gain  to  B's,  to  find  if  the  sum  equals  the  total  gain. 

RULE.  Write  one  partner's  investment  and  the  total  gain  as  a 
compound  numerator,  and  the  firm's  capital  as  the  denominator. 
Extend  the  resulting  fraction  if  necessary,  to  express  dollars  and 
cents. 

NOTE.  To  find  each  partner's  investment,  if  his  share  of  the  gain  is  given, 
write  one  partner's  gain  and  the  firm's  capital  as  a  compound  numerator,  and 
the  total  gain  as  the  denominator. 


PARTNERSHIP  257 

EXAMPLES  FOR   PRACTISE 

1.  A  and  B  united  in  conducting  a  grocery  business,  A  invest- 
ing $6480  and  B  $5670,  and  agreed  to  divide  the  gains  or  losses 
between  them  in  proportion  to  investment.     If  the  net  gains  were 
$3600,  how  much  should  each  partner  have  received? 

2.  A,  B,  and  C  have  formed  a  partnership,  A  investing  $8000, 
B  $10000  and  C  $12000.     The  net  gain  during  the  first  year  was 
$9000  which  was  divided  among  them  in  proportion  to  their 
respective  investments.     How  much  of  the  gain  should  each  part- 
ner have  received? 

3.  A  and  B  engaged  in  business  with  a  joint  capital  of  $23275; 
and  when  the  gains  were  adjusted,  A  received  $2400  and  B  $3600. 
How  much  did  each  invest?     [Note] 

4.  A,  B,  and  C  are  partners,  A  investing  $12000,  B  $15000 
and  C  $18000;   and  agreed  to  share  gains  or  losses  in  proportion 
to  their  investments.     When  the  books  were  adjusted,  it  was 
found  that  the  total  resources  of  the  firm  were  $78216  and  their 
total  liabilities  $24816.     What  was  each  partner's  share  of  the 
gain? 

DIVISION  OF  GAINS  OR  LOSSES  AMONG  PARTNERS  IN 
PROPORTION  TO  THEIR  AVERAGE  INVESTMENTS 

303.  INTRODUCTION,  (a)  To  divide  the  gains  or  losses  among 
partners  in  proportion  to  their  original  investments  as  in  302  is 
proper  only  when  those  investments  remain  unchanged  during 
the  entire  settlement-period.  If  a  change  is  made  in  the  original 
investment  of  one  or  more  partners  by  subsequent  investments 
or  withdrawals,  some  modification  of  302  will  be  necessary  to 
effect  an  equitable  distribution  of  gains  or  losses.  If,  for  instance, 
a  partner  were  to  make  an  investment  subsequent  to  his  original 
investment  and  thus  increase  his  contribution  to  the  gain-produc- 
ing capital  of  the  firm,  it  is  evident  that  he  should  receive  a 
corresponding  increase  of  the  firm's  gain  in  proportion  to  his  in- 
creased investment,  and  also  that  it  would  be  as  improper  to  pro- 
portion his  gains  for  the  entire  year  to  his  subsequent  greater 
investment  when  his  investment  for  the  preceding  portion  of  the 
year  was  less,  as  it  would  have  been  to  proportion  his  gains,  to  his 
original  less  investment  when  his  investment  for  a  subsequent 
portion  of  the  year  was  greater. 

(b)  This  modification  of  any  partner's  share  of  the  gain  in 


258  PARTNERSHIP 

proportion  to  his  modification  of  his  original  contribution  to  the 
profit  producing  capital  of  the  firm  is  equitably  effected  by  reduc- 
ing the  several  investments  of  each  partner  for  dissimilar  periods 
of  a  year  to  equivalent  investments  for  similar  periods  of  a  lower 
unit  of  time  than  a  year,  as  for  one  month,  or  for  one  day.  Hence, 
(c)  If  all  the  partners'  investments  were  not  made  for  the 
entire  settlement  period,  they  should  be  reduced,  by  average,  to 
equivalent  investments  for  the  longest  obtainable  equal  period  of 
time,  as  for  one  month  or  for  one  day;  and  the  gains  or  losses  should 
be  divided  among  the  partners  in  proportion  to  their  equivalent 
investments  for  that  common  lower  unit  of  time. 

ILLUSTRATIVE  EXAMPLE 

A  and  B  formed  a  copartnership,  and  agreed  to  share  gains  or 
losses  in  proportion  to  investment.  A  invested  $9000  on  Jan.  1, 
1910;  withdrew  $2000  on  May  1,  1910;  and  invested  $5000  on 
Aug.  1,  1910.  B  invested  $8000  on  Jan.  1,  1910;  invested  $4000 
on  Apr.  1,  1910;  and  withdrew  $3000  on  Sept.  1,  1910.  If  the  net 
gains  during  the  year  were  $6000,  what  was  each  partner's  share? 

SOLUTION 

$9000  invested  for  1  yr.    =12  times  $9000,  or  $108000  for  1  mo. 

5000  invested  for  5  mo.  =    5  times    5000,  or      25000  for  1  mo. 

A's  total  investment,  $133000  for  1  mo. 

$2000  withdrawn  for  8  mo.  =  8  times  $2000,  or      16000  for  1  mo. 

A's  net  investment,  $117000  for  1  mo. 

$8000,invested  for  1  yr.    =12  times  $8000,  or  $  96000  for  1  mo. 
4000  invested  for  9  mo.  =    9  times    4000,  or      36000  for  1  mo. 

B's  total  investment,  $132000  for  1  mo. 

$3000  withdrawn  for  4  mo.  =  4  times  $3000,  or      12000  for  1  mo. 

B's  net  investment,  $120000  for  1  mo. 

$117000  +  $120000  =  $237000,  firm's  net  capital  for  1  mo. 

39 
$6000  xWm       S234000       _  Q3  _   ^  ^  ^  ^  ^ 


79 

79 

40 

$6000  X  W9W       $240000 


79  $3037.97  +,  B's  share  of  the  gain 


79 


PARTNERSHIP  259 

EXPLANATION.  As  the  investments  of  the  partners  were  not  uniform 
throughout  the  year,  recourse  must  be  had  to  a  shorter  common  unit  of  time 
to  which  the  net  investments  of  the  several  partners  can  be  reduced,  and  in 
proportion  to  which  the  net  gain  of  the  firm  can  be  divided.  As  the  original 
investments  and  subsequent  modifications  were  all  made  on  the  first  day  of 
some  calendar  month,  the  investments  and  withdrawals  of  each  partner  are 
capable  of  being  reduced  by  average  to  equivalent  investments  and  with- 
drawals for  the  similar  period  of  one  month,  that  is,  in  A's  case  to  a  net  invest- 
ment of  $117000  for  1  mo.,  in  B's  case  to  a  net  investment  of  $120000  for  1  mo., 
and  with  reference  to  the  firm  to  an  equivalent  net  capital  of  $117000  + 
$120000,  or  $237000  for  1  mo.  Hence,  divide  the  net  gain  of  the  firm  ($6000) 
between  A  and  B  in  proportion  to  these  equivalent  investments  per  month, 
as  in  302  for  investments  per  year. 

RULE.  1.  //  all  the  investments  and  withdrawals  of  all  the  part- 
ners have  been  made  on  the  first  day  of  some  calendar  month,  divide 
the  gain  or  loss  of  the  firm  among  the  partners  as  in  302,  but  in  pro- 
portion to  each  partner's  average  net  investment  and  to  the  firm's 
average  net  capital  for  one  month. 

2.  //  one  or  more  investments  or  withdrawals  of  one  or  more  part- 
ners have  been  made  on  different  dates  of  a  calendar  month,  divide 
the  gain  or  loss  of  the  firm  among  the  partners  as  in  302,  but  in  pro- 
portion to  each  partner's  average  net  investment  and  to  the  firm's 
average  net  capital  for  one  day. 

NOTE  1.  To  find  a  partner's  average  net  investment  for  one  month  or 
for  one  day,  multiply  each  investment  or  withdrawal  by  the  number  of  months 
or  days  from  its  date  to  the  end  of  the  settlement-period,  and  subtract 
the  sum  of  the  resulting  withdrawal  products  from  the  sum  of  the  resulting 
investment  products. 

NOTE  2.  If  an  allowance  is  made  to  a  partner  for  special  labor,  or  risk, 
or  skill,  or  influence,  or  responsibility,  it  should  be  considered  as  an  extra  share 
of  the  gain  allotted  to  that  partner  for  that  special  investment,  in  addition 
to  his  proportionate  share  of  the  remainder  of  the  gain  which  is  divided  among 
all  the  partners  in  proportion  to  their  respective  investments  in  cash  or  its 
equivalent.  This  special  share  of  the  gain  should  therefore  be  deducted  from 
the  given  net  gain  (except  in  the  unusual  case  of  the  special  allowance  having 
been  previously  drawn  and  entered  in  the  expense  account),  before  such 
gain  is  proportioned  among  all  the  partners;  and  this  special  allowance  to 
a  partner  should  be  added  to  his  proportional  share  to  find  that  partner's 
actual  share  of  the  total  gain. 

NOTE  3.  In  applying  Note  1,  much  labor  will  be  saved  by  obtaining  the 
equivalent  net  investments  in  hundreds  of  dollars  for  one  month  or  for  one  day, 
as  shown  and  explained  in  the  abridged  solution  of  293. 


260  PARTNERSHIP 

EXAMPLES  FOR  PRACTISE 

1.  A  and  B  form  a  partnership  on  Jan.  1,  1910,  A  investing 
$9000  and  B  $6000.     On  June  1,  A  invested  $3000  additional,  and 
on  Sept.  1,  he  withdrew  $2000.     On  May  1,  B  withdrew  $2000, 
and  on  Aug.  1,  he  invested  $5000.     On  Jan.  1,  1911,  the  net  gains 
were  $4000,  which  were  divided  between  the  partners  hi  propor- 
tion to  their  respective  investments.     What  was  each  partner's 
share? 

2.  A,  B,  and  C  engaged  hi  partnership,  and  agreed  to  share 
gains  or  losses  hi  proportion  to  average  investment  per  month. 
A  invested  $12500  on  Jan.  1, 1911;  and  withdrew  $2500  on  June  1, 
1911.    B  invested  $9000  on  Jan.  1,  1911;  invested  $2000  on  Apr. 
1,  1911;  and  withdrew  $3000  on  Aug.  1,  1911.     C  invested  $15000 
on  Jan.  1,  1911,  and  permitted  it  to  remain  unchanged  throughout 
the  year.     If  the  net  gains  for  the  year  were  $7000,  what  was  each 
partner's  share?    What  was  each  partner's  interest  hi  the  firm 
after  the  adjustment? 

3.  A  and  B  engage  in  business  as  partners  on  Jan.  1,  1912, 
A  in  vesting  $15000  and  B  $18000;  and  agreed  to  share  gains  or 
losses  hi  proportion  to  average  investment  per  month.     A  made 
a  further  hi  vestment  of  $5000  on  Aug.  1,  1912;  and  B  withdrew 
$2000  on  Mch.  1,  1912.     On  May  1,  1912,  the  firm  took  in  C  as 
a  third  partner  who  then  hi  vested  $8000,  and  on  July  1,  1912, 
invested  an  additional  $6000.     On  Jan.  1,  1913,  the  capital  of  the 
firm  was  $65000.     What  was  each  partner's  interest  hi  the  firm 
on  the  latter  date? 

4.  A  and  B  are  partners,  having  agreed  to  share  gains  or 
losses  in  proportion  to  average  investment  per  day.     A  invested 
$18000  on  Jan.  1,  1911;  withdrew  $2000  on  May  16,  1911;   and 
withdrew  $1000  on  Sept.  12,  1911.     B  invested  $12000  on  Jan.  1, 
1911;    and  invested  $9000  on  July  23,  1911.     If  the  net  gains 
during  the  year  were  $6000,  what  was  each  partner's  share?    What 
was  each  partner's  interest  in  the  firm  on  Jan.  1,  1912? 

5.  A  and  B  united  in  partnership  and  agreed  to  share  gains 
or  losses  in  proportion  to  average  investment  per  day.     A  invested 
$8000  on  Jan.  1,  1911;   invested  $4000  on  May  19,  1911;    and 
withdrew  $2000  on  Oct.  8;  1911.    B  invested  $12000  on  Jan.  1, 


PARTNERSHIP  261 

1911;  withdrew  $2000  on  Apr.  23,  1911;  and  withdrew  $1000  on 
Nov.  2,  1911.  What  was  each  partner's  interest  in  the  firm  on 
Jan.  1,  1912,  if  its  capital  at  that  time  was  $25000? 

6.  A  and  B  engaged  in  business  as  partners  and  agreed  to 
share  gains  or  losses  in  proportion  to  average  investment  per  day. 
A  invested  $12000  on  Jan.  1,  1911;  invested  $7000  on  June  20, 
1911;  and  withdrew  $4000  on  Oct.  12,  1911.  B  invested  $15000 
on  Jan.  1,  1911;  withdrew  $3000  on  Apr.  29,  1911;  and  withdrew 
$3000  on  Aug.  24,  1911.  On  Jan.  1,  1912,  the  books  were  adjusted, 
when  it  was  found  that  the  total  resources  of  the  firm  were  $40000, 
and  their  total  liabilities  $8000.  What  was  each  partner's 
interest  in  the  firm  on  Jan.  1,  1912? 

WHEN  EACH  PARTNER'S  PROPORTIONATE  PART  OF  THE 
GAIN  OR  LOSS  IS  ARBITRARILY  FIXED  IN  THE  PARTNER- 
SHIP AGREEMENT,  AND  HIS  INVESTMENT  IS  REQUIRED 
TO  BE  IN  PROPORTION  TO  HIS  STIPULATED  SHARE  OF 
THE  GAINS  OR  LOSSES 

304.  INTRODUCTION,  (a)  A  third  method  of  distributing 
gains  or  losses  among  partners,  and  in  many  respects  the  most 
equitable,  is  to  assign  to  each  partner  a  fixed  part  of  the  gain, 
irrespective  of  his  pecuniary  investment.  There  are  many  pos- 
sible forms  of  investment  besides  that  of  money.  Any  contribu- 
tion of  a  partner  to  a  business  which  will  add  materially  to  its 
prosperity,  such  as  superior  business  talent,  greater  social  influ- 
ence, excess  of  time  or  labor,  disproportionate  responsibility  in 
management,  etc.,  are  properly  investments,  and  frequently  the 
most  important  elements  in  acquiring  gains.  If  one  partner 
invests  twice  as  much  money  as  another  partner,  that  of  itself 
should  not  be  a  sufficient  reason  for  giving  him  twice  as  much  of 
the  gain,  except  he  also  invest  twice  as  much  time,  twice  as  much 
responsibility  of  management,  and,  generally,  twice  as  much  of 
all  the  contributory  elements  in  the  acquisition  of  the  firm's  gains. 

(6)  A  mere  difference  in  pecuniary  investment  can  readily 
be  adjusted  between  two  partners  if  they  are  equal  in  all  other 
respects,  by  giving  to  each  partner  one-half  the  firm's  net  gains, 
and  requiring  the  partner  who  invested  less  than  one-half  the  cap- 
ital to  pay  interest  upon  his  deficiency  at  the  market  rate  per 
annum,  to  the  other  partner  who  must  have  invested  correspond- 
ingly more  than  one-half  the  capital,  since  both  together  must 
have  contributed  two-halves  or  the  entire  capital. 

(c)  If,  however,  the  partners  are  unequal  in  other  respects 


262  PARTNERSHIP 

than  the  money  which  they  have  invested,  that  is,  if  A  of  the  firm 
of  A  and  B,  is,  in  all  those  qualities  which  put  together  consti- 
tute a  successful  business  man,  twice  as  capable  as  B,  but  finan- 
cially is  deficient,  then  A  should  equitably  receive  twice  as  much 
of  the  firm's  net  gains  as  B,  or  two-thirds  to  B's  one-third;  and 
as  the  proper  measure  of  the  use  of  money  is  the  interest  thereon 
at  the  current  rate  per  annum  for  the  time  it  is  used,  A  should 
pav  to  B  interest  on  any  deficiency  of  his  investment  below  two- 
thirds  of  the  entire  capital.  Conversely  B  should  receive  one- 
third  of  the  firm's  net  gain  and  receive  interest  on  the  excess  of 
his  financial  investment  above  one-third  of  the  entire  capital; 
for  B's  excess  of  investment  above  one-third  of  the  firm's  capital 
must  exactly  balance  A's  deficiency  below  two-thirds  of  the  firm's 
capital,  as  the  firm's  capital,  or  what  both  have  invested,  must 
equal  three-thirds  of  itself. 

(d)  The  allowance  of  interest  by  one  partner  who  is  deficient 
in  his  required  average  investment  to  another  partner  who  is  in 
excess,  is  most  conveniently  accomplished  by  crediting  each  part- 
ner with  the  interest  on  his  net  average  investment;  and  debiting 
the  firm's  interest  account,  or  its  practical  equivalent  of  subtract- 
ing the  total  interest  on  the  several  partners'  net  average  investments 
from  the  firm's  net  gains;  and  proportioning  the  remainder  of  the 
net  gain  among  the  partners  in  accordance  with  the  partnership 
agreement. 

ILLUSTRATIVE  EXAMPLE 

A  and  B  engaged  in  business  as  partners  on  Jan.  1,  1911,  the 
agreement  between  them  specifying  that  A,  by  reason  of  his  su- 
perior business  ability,  should  receive  -f-  of  the  net  gain  of  the 
business,  and  B  the  remaining  f;  and  that  each  should  be  account- 
able for  investing  as  many  sevenths  of  the  net  capital  as  he  was  to 
receive  sevenths  of  the  net  gain,  and  be  required  to  pay  interest 
at  6  %  upon  as  much  of  his  required  number  of  sevenths  of  the  net 
capital  as  he  failed  to  invest,  and  be  entitled  to  receive  interest 
at  the  same  rate  per  annum  on  his  excess  if  he  invested  more  than 
his  stipulated  number  of  sevenths.  At  the  end  of  the  year  the 
net  gain  was  found  to  be  $4291.  A  invested  $12000  on  Jan.  1, 
1911;  withdrew  $4000  on  Apr.  16,  1911;  invested  $7000  on  Aug. 
25,  1911;  and  withdrew  $3000  on  Oct.  8,  1911.  B  invested  $15000 
on  Jan.  1,  1911;  withdrew  $2000  on  Mch.  17,  1911;  withdrew 
$1500  on  July  5,  1911;  and  invested  $5000  on  Sept.  3,  1911. 
What  was  each  partner's  interest  in  the  firm  at  the  end  of  the  year? 


PARTNERSHIP  263 

SOLUTION 

Int.  of  $12000  from  Jan.  1  to  Dec.  31,  inclusive  (1  yr.)      $720. 
Int.  of  $7000  from  Aug.  25  to  Dec.  31,  inclusive  (128  da.)     149.33+ 
Interest  on  A's  total  investments  $869.33 

Int.  of  $4000  from  Apr.  16  to  Dec.  31,  inclusive  (259  da.)  $172.67- 
Int.  of  $3000  from  Oct.  8  to  Dec.  31,  inclusive  (84  da.)        42. 

Interest  on  A's  total  withdrawals  $214.67 

$869.33  -  $214.67  =  $654.66,  int.  on  A's  net  average  investment. 

Int.  of  $15000  from  Jan.  1  to  Dec.  31,  inclusive  (1  yr.)   $900. 
Int.  of  $5000  from  Sept.  3  to  Dec.  31,  inclusive  (119  da.)      99.17- 
Interest  on  B's  total  investments  $999.17 

Int.  of  $2000  from  Mch.  17  to  Dec.  31,  inclusive  (289  da.)  $  96.33+ 
Int.  of  $1500  from  July  5  to  Dec.  31,  inclusive  (179  da.)     44.75 
Interest  on  B's  total  withdrawals  $141.08 

$999.17  -  $141.08  =  $858.09,  int.  on  B's  net  average  investment. 

$654.66,  obtained  int.  on  A's  net  average  investment. 
858.09,  obtained  int.  on  B's  net  average  investment. 
$1512.75,  interest  on  firm's  net  average  capital. 

Firm's  net  gain  ($4291)  —  interest  on  firm's  net  average  capi- 
tal ($1512.75)  =  firm's  gain  to  be  proportioned  among  the  part- 
ners as  per  agreement  ($2778.25).  [See  Introduction,  d.] 

$  of  $2778.25  =  $1587.57  (+  A's  int.  on  net  av.  inv.)  =  A's  gain 
f  of  $2778.25  =  $1190.68  (+  B's  int.  on  net  av.  inv.)  =  B's  gain 

($12000  +  $7000)  -  ($4000  +  $3000),  or  A's  net  in- 
vestment =  $12000. 
A's  stipulated  proportion  of  the  firm's  net  gain     =      1587.57 
A's  interest  on  net  average  investment  =        654.66 
A's  interest  in  the  firm  at  end  of  year  =  $14242.23 

($15000  +  $5000)  -  ($2000  +  $1500),  or  B's  net  in- 
vestment =  $16500. 
B's  stipulated  proportion  of  the  firm's  net  gam  =      1190.68 
B's  interest  on  net  average  investment  858.09 
B's  interest  hi  the  firm  at  end  of  year  =  $18548.77 


264  PARTNERSHIP 

EXPLANATION.  Compute  the  interest  upon  each  of  A's  investments  and 
withdrawals  from  its  date  to  the  end  of  the  settlement  period  (Dec.  31,  1911, 
inclusive),  and  subtract  the  total  interest  on  his  withdrawals  from  the  total 
interest  on  his  investments  to  find  the  total  interest  on  A's  net  average 
investment  ($654.66),  with  which  A  should  be  credited. 

Similarly  find  the  interest  on  B's  net  average  investment,  obtaining 
$858.09,  with  which  B  should  be  credited. 

Deduct  A's  and  B's  special  allowance  for  interest  ($654.66  +  $858.09  = 
$1512.75)  from  the  net  gain  of  the  firm  ($4291)  to  find  the  remainder  of  the 
net  gain  ($2778.25)  which  is  to  be  divided  between  the  partners  as  per  agree- 
ment, f  of  $2778.25  ($1587.57)  being  credited  to  A,  and  \  of  $2778.25  ($1190.68) 
being  credited  to  B. 

The  sum  of  either  partner's  net  investment,  the  interest  on  his  net  average 
investment,  and  his  stipulated  proportion  of  the  net  gain  after  the  net  gain 
had  been  diminished  by  the  interest  on  the  firm's  net  average  capital,  will  be 
that  partner's  interest  in  the  firm  at  the  time  of  adjustment. 

NOTE  1.  If  the  interest  on  any  partner's  total  investments  is  greater 
than  the  interest  on  his  total  withdrawals,  the  difference  between  these  totals 
will  be  the  interest  on  that  partner's  net  average  investment  for  which  he 
should  be  credited;  but  if  less,  the  difference  between  these  totals  will  be  the 
interest  on  that  partner's  net  average  insolvency  for  which  he  should  be  debited. 

NOTE  2.  If  the  average  financial  condition  of  each  of  the  partners  for 
the  settlement  period,  without  an  exception,  is  solvent,  the  sum  of  the  interest 
on  their  respective  net  average  investments  will  be  the  interest  on  the  firm's 
net  average  capital  for  the  same  period;  and  if  insolvent,  the  sum  of  the 
interest  on  their  respective  net  average  insolvencies  will  be  the  interest  on 
the  firm's  net  average  insolvency.  If,  however,  the  average  financial  condi- 
tion of  one  or  more  of  the  partners  is  solvent  and  that  of  others  insolvent,  the 
difference  between  the  ftum  of  all  the  solvent  partners'  interest  on  their  net 
average  investments  and  the  sum  of  all  the  insolvent  partners'  interest  on 
their  net  average  insolvencies,  will  be  the  interest  on  the  firm's  net  average 
capital  if  the  sum  of  the  interest  on  the  net  average  solvencies  is  the  greater, 
or  it  will  be  the  interest  on  the  firm's  net  average  insolvency  if  the  sum  of  the 
interest  on  the  net  average  insolvencies  is  the  greater. 

NOTE  3.  Subtract  the  interest  on  the  firm's  net  average  capital  from  the 
net  gain,  or  add  it  to  the  net  loss,  to  find  the  gain  or  loss  which  is  to  be  appor- 
tioned among  the  partners  in  accordance  with  the  partnership  agreement, 
for  if  the  firm's  "interest  account"  had  been  debited  for  the  interest  on  each 
solvent  partner's  net  average  investment  for  which  that  partner  had  been 
credited  and  that  "interest  account"  had  been  credited  with  the  interest 
on  each  insolvent  partner's  net  average  insolvency  for  which  that  partner 
had  been  debited,  and  the  firm's  "interest  account"  had  been  balanced, 
and  the  obtained  balance  had  been  included  in  finding  the  firm's  gain  or 
loss,  it  would  have  correspondingly  diminished  the  gain  or  increased  the  loss. 


PARTNERSHIP  265 

NOTE  4.  Add  the  interest  on  the  firm's  net  average  insolvency  to  the 
firm's  net  gain,  or  subtract  it  from  the  firm's  net  loss,  to  find  the  gain  or  loss 
which  is  to  be  apportioned  among  its  several  partners;  for  this  interest  ex- 
presses the  net  debit  interest  of  the  several  partners  for  which  the  firm's 
"interest  account"  should  exhibit  a  corresponding  "credit  balance."  If  this 
"credit  balance"  had  been  included  in  finding  the  firm's  gain  or  loss,  it  would 
have  correspondingly  increased  the  firm's  net  gain  or  diminished  the  firm's 
net  loss. 

EXAMPLES  FOR  PRACTISE 

1.  A  and  B  engaged  in  business  as  partners  on  Jan.  1,  1911, 
A  agreeing  to  invest  |  of  the  capital  and  to  receive  f  of  the  net 
gain  of  the  firm,  and  B  to  invest  $  of  the  capital  and  to  receive  £ 
of  the  net  gain.     It  was  further  agreed  that  interest  at  8  %  annum 
would  be  allowed  to  either  partner  for  investing  more  than  his 
stipulated  number  of  fifths  of  the  capital,  and  to  be  charged  against 
any  partner  for  any  deficiency  of  stipulated  investment.     A  in- 
vested $9000  on  Jan.  1,  1911;   invested  $3000  on  Apr.  12,  1911; 
withdrew  $1500  on  June  9,  1911;  invested  $2500  on  Aug.  13,  1911; 
and  withdrew  $900  on  Nov.  8,   1911.     B's  net  investment  was 
$18000,   and  the   interest  on  his  net   average  investment  was 
$758.90.     If  the  net  gain  of  the  firm  during  the  year  was  $7500, 
what  was  each  partner's  interest  hi  the  business  on  Jan.  1,  1912 
(that  is,  on  Dec.  31,  1911,  inclusive)? 

2.  A  and  B  formed  a  partnership  on  Jan.  1,  1912,  and  agreed 
to  invest  equally  and  to  share  equally  in  gains  or  losses,  and  in- 
terest at  9%  per  annum  was  to  be  allowed  to  either  partner  for 
any  excess  over  stipulated  investment,  or  to  be  charged  against 
either  partner  for  any  deficiency  of  investment.     The  net  gain  of 
the  firm  on  Jan.  1,  1913,  was  found  to  be  $7600.     A  invested 
$15200  on  Jan.  1,  1912;   withdrew  $2600  on  May  23,  1912;   and 
invested  $3800  on  Sept.  18,  1912.     B  invested  $18000  on  Jan.  1, 
1912;  invested  $1600  on  Mch.  17,  1912;  withdrew  $2000  on  June 
8,  1912;   and  invested  $3000  on  July  16,  1912.     What  was  each 
partner's  interest  in  the  firm  on  Jan.  1,  1913? 

3.  A,  B   and   C  entered  into  a  copartnership  agreement  to 
commence  business  on  Jan.  1,  1911,  and  agreed  to  invest  equally 
and  to  share  equally  in  gains  or  losses,  and  to  allow  interest  at 
6  %  per  annum  for  any  excess  over  stipulated  investment,  and  to 


266  PARTNERSHIP 

charge  interest  at  the  same  rate  for  any  deficiency.  A  invested 
$16000  on  Jan.  1,  1911;  invested  $4000  on  May  26,  1911;  and 
withdrew  $2400  on  July  12,  1911.  B  invested  $18600  on  Jan.  1, 
1911;  and  withdrew  $1500  on  Oct.  19,  1911.  C  invested  $14000 
on  Jan.  1,  1911;  withdrew  $2000  on  May  4,  1911;  and  invested 
$5000  on  Aug.  26,  1911.  If  the  net  gain  of  the  firm  at  the  end  of 
the  year  was  $12600,  what  was  each  partner's  interest  hi  the 
business  at  that  time? 

4.  A,  B  and  C  enter  into  partnership  to  engage  in  business,  A 
agreeing  to  invest  f  of  the  capital,  B  to  invest  f ,  and  C  f,  and  to 
receive  the  same  proportion  of  the  net  gain  of  the  business.  It 
was  also  agreed  that  each  partner  was  to  be  credited  with  interest 
at  10  %  per  annum  for  any  excess,  and  to  be  debited  with  interest 
at  the  same  rate  for  any  deficiency,  of  stipulated  investment.  A 
invested  $9000  on  Jan.  1,  1912;  and  withdrew  $500  on  June  25, 
1912.  B  invested  $14000  on  Jan.  1,  1912;  and  invested  $3000  on 
May  14,  1912.  C  invested  $7500  on  Jan.  1,  1912;  invested  $3500 
on  July  2,  1912;  and  withdrew  $2000  on  Sept.  20,  1912.  On  Jan. 
1,  1913,  the  books  were  adjusted  to  the  end  of  the  preceding  day, 
and  showed  a  net  gain  of  $8730.  What  was  each  partner's  in- 
terest hi  the  firm  at  settlement? 

6.  A  and  B  entered  into  partnership,  A  agreeing  to  invest  -ft 
of  the  capital  and  to  receive  TT  of  the  total  gain,  and  B  to  invest 
TT  of  the  capital  and  to  receive  TT  of  the  total  gain,  and  interest 
at  a  fixed  rate  per  annum  was  to  be  credited  to  either  partner  for 
any  excess,  or  debited  against  him  for  any  deficiency,  of  required 
investment.  At  settlement  it  was  discovered  that  the  net  capital 
of  the  firm  was  $26200;  that  A's  total  interest  on  investments 
was  $518.65,  his  total  interest  on  withdrawals  was  $168.15,  and  his 
net  hi  vestment  $15300;  and  that  B's  total  interest  on  invest- 
ments was  $927.83;  his  total  interest  on  withdrawals  $216.48;  and 
his  net  investment  $8700.  Find  each  partner's  interest  in  the 
firm  at  settlement. 


APPENDIX 
USEFUL  COMMERCIAL  CALCULATIONS 

THE  ALIQUOT  METHOD  OF  COMPUTATION 

305.  An  aliquot  of  a  number  is  any  number  which  will  exactly 
divide  it. 

NOTE.  An  aliquot  part  of  a  given  number  is  the  fractional  part  which 
expresses  the  relation  of  the  aliquot  to  that  number.  Thus,  8  is  an  aliquot  of 
32,  and  J  is  the  aliquot  part  which  8  is  of  32.  The  following  are  aliquots  of 
one  dollar  (100  cents)  : 


*} 

25*    =$i 


ILLUSTRATIVE  EXAMPLE 
306.   What  is  the  cost  of  52  yd.  muslin  at  12J^f  per  yard? 

SOLUTION.  \1\t  =  $J.  Therefore  at  $|  per  yd.,  8  yd.  will  cost  $1,  and 
52  yd.  will  cost  as  many  times  $1  as  8  yd.  are  contained  times  in  52  yd.,  or  6£ 
times  $1  =  $6.50. 

RULE.  Divide  the  given  quantity  by  the  quantity  which  $1  can 
buy. 

EXAMPLES  FOR  PRACTISE 
Find  the  cost  of 

1.  512yd.  at  12^  per  yd.  7.   292  gal.  at  6i  i  per  gal. 

2.  748  Ib.  at  25^  per  Ib.  8.   610  Ib.  at  5^  per  Ib. 

3.  465  gal.  at  20  i  per  gal.  9.   583  yd.  at  25  i  per  yd. 

4.  236  bu.  at  50^  per  bu.  10.   329  Ib.  at  Vl\i  per  Ib. 

5.  726  Ib.  at  16f  i  per  Ib.  11.   837  gal.  at  50?f  per  gal. 

6.  417  yd.  at  33j^  per  yd.  12.   913  doz.  at  20?f  per  doz. 


268  APPENDIX 

ILLUSTRATIVE  EXAMPLE 

307.  At  25*  per  pound,  how  many  pounds  of  cheese  can  be 
bought  for  $9? 

SOLUTION.  25£  =  $^.  Therefore  at  Sj  per  lb.,  $1  can  buy  4  lb.,  and 
9  times  $1,  or  $9,  can  buy  9  times  4  lb.,  or  36  lb. 

RULE.  Multiply  the  quantity  which  $1  can  purchase  by  the 
number  of  dollars  to  be  expended,  expressing  cents  in  the  total  cost  as 
a  fraction  of  a  dollar. 

EXAMPLES  FOR  PRACTISE 

How  much  can  be  bought  for 

1.  $375  at  20*  per  yd.?        5.  $61.50  at  25  *  per  gal.? 

2.  $291  at  16f*pergal.?      6.  $41.25  at  12£*  per  yd.? 

3.  $148  at  25*  per  lb.?         7.  $278.75  at  33  J*  per  lb.? 

4.  $86  at  50*  per  bu.?          8.  $109.20  at  45*  per  doz.? 

308.  A  price  which  is  not  itself  an  aliquot  of  $1  may  frequently 
be  resolved  into  two  or  more  components  which  are.     Thus, 


7}*  =  5*  +  2H  or  $2V  +  \  of 
15*  =  10*  +  5*,  or  SrV  +  }  of 
18f*  =  12}*  +  6K  or  $J  +  i  of  $J. 
22^  =  20^f  +  2H  or  $i  +  i  of  Si. 
27}*  =  25*  +  2}*,  or  Si  +  TV  of  Si- 
37^*  =  25*  +  12}*,  or  Si  +  }  of  Si- 
55*  =  50*  +  5*,  or  S}  +  TV  of  S}. 
62J*  =  50*  +  12}*,  or  $}  +  J  of  S}. 
75*  =  50*  +  25*  or  $J  +  }  of  S}. 

ILLUSTRATIVE  EXAMPLE 
309.    Find  the  cost  of  764  yd.  cloth  at  37}*  per  yard. 

SOLUTION 

At  25*    per  yd.,  764  yd.  will  cost  $191       (i  of  $764). 
«     «      «     "      «      «        05  *n  a  rkf  «i  cm 


EXAMPLES  FOR  PRACTISE 
Find  the  cost  of 

1.  618yd.  at  18f?fperyd.       5.   640  yd.  at  55 1  per  yd. 

2.  428  Ib.  at  15?f  per  Ib.  6.   732  gal.  at  22} fi  per  gal. 

3.  316  bu.  at  62}^  per  bu.       7.   684  Ib.  at  $1.37}  per  Ib. 

4.  520  gal.  at  7J?f  per  gal.        8.   562  bu.  at  $1.15  per  bu. 

NOTE  1.     The  above  process  can  be  advantageously  applied  to  common 
fractions  of  such  denominators  as  are  usually  employed  in  business. 

Multiply  (9)  $216  by  T5*;     (10)  $756  by  &. 

SOLUTION 

(9)  (10) 

$216  =  16  sixteenths  $756  =  32  thirty-seconds 

\  of  216  =  54    =  4  }  of  756  =  189  =    8       "          " 

}of    54=JL3}  =  1        "  j  of  189  =   23 f=    1       " 

$67}  or  $67.50  =  fV  $212f  or  $212.625  =  •&• 

EXAMPLES  FOR  PRACTISE 

Multiply                           Multiply  Multiply 

11.  $648  by  f          14.   824  Ib.  by  A        17.  372  gal.  by  & 

12.  $432  by  T3*        15.   268  yd.  by  «       18.  716  yd.  by  « 

13.  $596  by  A        16.   562  bu.  by  \\       19.  384  bu.  by  If 

310.   A  price  which  is  not  itself  an  aliquot  of  $1,  may  lack  an 
aliquot  of  being  $1.     Thus, 

99^     =  $1  - -riv  of  $1.        95^f     =  $1  -  ^  of  $1 
$l-*Vof$l  87}^f  =  $1  -  iof  $1 

$1  -A  of  $1  75^    =$l-iof$l 

ILLUSTRATIVE  EXAMPLE 
Find  the  cost  of  136  yd.  cloth  at  75  f  per  yd. 

SOLUTION 

136  yd.  at  $1  per  yd.  =  $136. 
"     "     "  $J    "     «    =      34  Q  of  $136). 
"     "     "  $J    "     "     =  $102  ($136  -  $34). 


270  APPENDIX 

EXAMPLES  FOR  PRACTISE 
Find  the  cost  of 

1.  260  yd.  @  95^        4.   473  gal.  @  99^f         7.    152  bu.  @  $1.95 

2.  512  Ib.  @  87 \i       5.   390  bu.  @  98^          8.   284  bbl.  @  $6.90 

3.  432  gal.  @  75^       6.   256  bbl.  @  $5.75       9.   520  bbl.  @  $7.80 

NOTE  1.  The  above  process  can  also  be  used  when  the  multiplier  is  a 
common  fraction  and  its  numerator  lacks  an  aliquot  of  being  equal  to  its 
denominator. 

Multiply  (10)  1264  by  f ;        (11)  928  by  }. 

SOLUTION 
(10)  (11) 

1264  =  8  eighths  928  =  4  fourths 

\  of  1264  =    158  =  1       "  i  of  928  =  232  =  1 

(1264  -  158)  =  1106  =  |  (928  -  232)  =  696  =  f 

EXAMPLES  FOR  PRACTISE 
Multiply  Multiply  Multiply 

12.  519  by  |  15.   453  by  £  18.  293  by  4J 

13.  673  by  f  16.   369  by  |  19.  467  by  2| 

14.  527  by  |  17.   258  by  A  20.  625  by  3| 

NOTE  2.  The  above  process  will  also  be  found  useful  when  a  multiplier 
lacks  12  or  less  of  being  100,  1000,  10000,  etc. 

21.  Multiply  386  by  997. 

SOLUTION  EXPLANATION.     For  convenience,  first  mul- 

386  X  1000  =  386000       ^p*y  ^y  1000,  as  explained  in  30,  obtaining  a 
38fi  v  o  1 1  co       result  which  is  3  times  386  too  great.     Hence, 

deduct  3  times  386  from  the  result,  obtaining 
386  X  997     =  384842       384842  as  the  correct  product. 

EXAMPLES  FOR  PRACTISE 
Multiply  Multiply  Multiply 

22.  875  by  97  25.     52  by  995  28.  712    by  9998 

23.  426  by  99  26.   628  by  994  29.  256    by  980 

24.  642  by  96  27.   329  by  993  30.  7384  by  9200 


TO  FIND  THE  COST  OF  GRAIN  WHEN  THE  PRICE  IS  PER 

BUSHEL  AND  THE  QUANTITY  OF  GRAIN  IS  EXPRESSED 

IN  POUNDS 

ILLUSTRATIVE  EXAMPLE 
311.   Find  the  cost  of  5687  Ib.  oats  at  49  £  per  bushel  of  32  Ibs. 

SOLUTION 

5687 

$.49  EXPLANATION.     At    49?f    per    pound,    the    cost    would 

5H83  have  been  5687  times  49^,  or  $2786.63;   therefore  at  49?$ 

_  per  bushel  of  32  pounds,  this  obtained  cost   must    be  32 

^*    °  times  the  correct    cost.     Hence,  divide  $2786.63  by  32,  or 

8)2786.63  by  the  factors  of  32  (8  and  4)  to  find  the  correct  cost. 

4)348.329- 
$87.082 

RULE.  Multiply  the  weight  in  pounds  by  the  price  per  bushel, 
and  divide  the  product  by  the  weight  in  pounds  of  1  bushel  of  the  given 
kind  of  grain. 

EXAMPLES  FOR  PRACTISE 

Find  the  cost  of 

1.  15207  Ib.  wheat  at  $1.09  per  bu.  of  60  Ib.  (10  X  6). 

2.  9873  "  corn     "     68#    "     "    "  56  "   (8  X  7). 

3.  5928  "  beans    "  $2.10   "     "    "    60  " 

4.  1753  "  timothy  seed  at  $2.35  per  bu.  of  45  Ib.  (9  X  5). 

5.  2874  "  buckwheat  at  $1.25  per  bu.  of  48  Ib.  (8  X  6). 

NOTE.  The  above  process  can  be  advantageously  employed  in  finding 
the  cost  of  any  merchandise  when  the  quantity  is  expressed  in  units  of  one 
denomination  and  the  price  is  per  unit  of  a  different  denomination,  thus  avoid- 
ing an  awkward  intermediate  fraction. 

6.  5368  Ib.  coal  at  $6.75  per  ton  of  2000  Ib.  [39]. 

7.  18249   "      "     "  $7.25    "      "    "      "      " 

8.  27235  "      "     "  $6.50    "      "    "      "      " 

9.  8739  "     "     "$7.10    "     "    "2240  "(10X8X7X4). 

10.  14296   "     "     "  $7.50    "      "    "  2240    " 

11.  3876  "  ear  corn  at  $3.25  per  bbl.  of  350  Ib.  (10  X  7  X  5). 

12.  9285  "  freight  at  45 £  per  cwt.  [38]. 


272 


APPENDIX 


13.  25462  ft.  lumber  at  $18.50  per  M. 

14.  12348  Ib.  dressed  beef  at  $10.15  per  100  Ib. 

15.  6285  Ib.  flour  at  $5.20  per  bbl.  of  196  Ib.  (7  X  7  X  4). 


ADDITION  BY  GROUPING  FIGURES 

312.  Skill  in  addition  is  an  invaluable  accomplishment  which 
should  be  acquired  by  every  one  who  wishes  to  follow  a  business 
career.  No  method  of  addition  has  so  many  advantages  and  so 
few  disadvantages  as  that  of  adding  one  column  at  a  time  and, 
while  doing  so,  training  the  eye  to  take  within  the  range  of  its 
vision  as  many  successive  figures  in  the  column  as  the  mind  may 
be  capable  of  simultaneously  adding.  At  first,  the  learner  should 
attempt  to  group  only  those  figures  which  are  within  the  power  of 
his  present  capacity,  and  let  actual  experience  demonstrate  the 
limitations  of  his  individual  ability.  Efficiency  in  grouping  will 
increase  by  exercise.  10  is  the  usual  maximum  grouping  recom- 
mended for  beginners. 


!}• 

2) 

3^-9 

4) 

9 

2 

6V-9 


10 


10 


6) 

1  ^10 
3) 


1) 
4J- 

3) 


8 


4 
3 
2 

61 
3) 

9 


10 


2 


10 


82 


ILLUSTRATIVE  EXAMPLE 


EXPLANATION.  Successive  figures  in  the 
several  columns  which  admit  of  grouping  into 
convenient  combinations  are  here  connected  by 
a  brace  to  enable  the  learner  the  better  to  com- 
prehend the  manner  in  which  figures  may  be 
thus  massed.  The  higher  figures  as  9's  or  8's, 
which  have  no  correspondingly  low  contiguous 
figures,  as  1's  or  2's,  that  can  be  combined  with 
them,  should  be  added  singly. 

Thus,  add  the  groupings  of  the  units'  column 
(8,  17,  27,  33,  42,  52,  62,  69),  obtaining  69. 
Write  9  units  in  the  units'  column,  and  carry 
6  tens  to  the  tens'  column. 

6  (to  carry),  15,  24,  34,  43,  53,  61,  70,  80, 
obtaining  80  tens,  or  8  hundreds  and  0  tens. 
Write  0  tens  in  the  tens'  column,  and  carry  8 
hundreds  to  the  hundreds'  column. 

8  (to  carry),  16,  26,  36,  46,  55,  64,  73,  82, 
obtaining  82  hundreds,  making  the  sum  of  the 
three  columns  equal  8209. 


},. 


ADDITION   BY    GROUPING    FIGURES 


273 


EXAMPLES  FOR 

PRACTISE 

(1) 

(2) 

(3) 

(4) 

(6) 

(6) 

(7) 

3 

2 

21 

456 

142 

326 

569 

5 

7 

62 

633 

357 

762 

413 

4 

1 

93 

291 

213 

691 

324 

9 

4 

35 

832 

431 

325 

741 

6 

3 

21 

165 

565 

443 

246 

2 

2 

58 

312 

252 

232 

632 

8 

5 

14 

421 

343 

586 

171 

7 

4 

63 

256 

224 

213 

514 

2 

9 

31 

432 

769 

732 

465 

6 

6 

58 

318 

132 

964 

922 

4 

2 

42 

271 

348 

143 

873 

5 

3 

75 

726 

622 

212 

221 

3 

7 

34 

193 

896 

747 

512 

(8) 

(9) 

(10) 

(ID 

(12) 

2674 

64287 

487639 

528964 

8492716 

3425 

13612 

261324 

272625 

1637154 

4283 

32291 

432152 

313428 

1231142 

1716 

48453 

145464 

293112 

3536126 

9134 

21432 

925515 

654577 

4172641 

2942 

32125 

343123 

132322 

2629239 

7123 

76562 

326682 

721831 

7314275 

1358 

23128 

471307 

276146 

2163913 

2542 

94319 

819731 

819120 

1342592 

3291 

62643 

961252 

125394 

4289411 

4876 

24356 

327218 

363283 

4526956 

9212 

25191 

712892 

142317 

2133242 

2794 

43737 

231429 

654976 

3641727 

6243 

12453 

546561 

211732 

5324361 

2922 

56214 

324326 

732142 

2472613 

4636 

31925 

453656 

129124 

1413486 

1256 

93172 

652146 

476164 

6235612 

5126 

59213 

254916 

323622 

2249232 

1716 

29561 

756926 

342429 

7264145 

8226 

39343 

154976 

248928 

2295724 

274  APPENDIX 


CROSS-MULTIPLICATION 

313.  In  obtaining  the  necessary  extensions  of  many  of  the 
separate  items  in  making  out  bills,  invoices,  etc.,  much  valuable 
time  will  be  saved  by  the  following  process,  when  the  price  and 
the  quantity  consist  of  2  or  3  orders,  as  is  ordinarily  the  case. 

ILLUSTRATIVE  EXAMPLE 
Find  the  cost  of  23  yards  cloth  at  45  cents  per  yard. 

SOLUTION         EXPLANATION.     The  product  of  two  numbers  is  the  product 
A  .  -         of  each  order  of  one  number  separately  multiplied  by  each  order 
of  the  other  number  [Prin.  2,  26].     Hence, 

First  multiply  the  units'  orders  (3  times  5  units  =  15  units, 


$10.35  or  1  ten  and  5  units).  Write  5  units  as  the  units'  order  of  the 
product,  and  carry  1  ten. 

Second,  multiply  each  tens'  order  by  its  opposite  units'  order  (3  times  4 
tens  +  5  times  2  tens  +  1  ten  to  carry  =  23  tens,  or  2  hundreds  and  3  tens). 
Write  3  tens  as  the  tens'  order  of  the  product,  and  carry  2  hundreds. 

Third,  multiply  the  tens'  orders  (4  tens  X  2  tens  =  8  hundreds)  +  2 
carried  hundreds  =  10  hundreds;  thus  making  the  complete  product  1035 
cents,  or  $10.35. 

EXAMPLES  FOR  PRACTISE 


Multiply 

Multiply 

Multiply 

Multiply 

1. 

48  by 

37 

4. 

86  by 

63 

7. 

27  by 

35 

10. 

73 

by  54 

2. 

53  by 

74 

5. 

58  by 

24 

8. 

65  by 

23 

11. 

36 

by  27 

3. 

29  by 

82 

6. 

95  by 

32 

9. 

42  by 

67 

12. 

52 

by  73 

POWERS  AND  ROOTS 

314.  A  power  of  a  number  is  the  product  of  that  number  if 
taken  two  or  more  times  as  a  factor. 

NOTE  1.  The  second  power  or  square  of  3  is  the  product  of  3  taken  twice 
as  a  factor  (3  X  3),  or  9;  the  third  power  or  cube  of  2  is  the  product  of  2  taken 
three  times  as  a  factor  (2  X  2  X  2),  or  8;  the  fourth  power  of  5  is  the  product 
of  5  taken  four  times  as  a  factor  (5  X  5  X  5  X  5),  or  625;  etc. 

NOTE  2.  The  exponent  of  a  power  is  a  small  figure  placed  diagonally 
above  a  number  to  denote  how  many  times  it  is  to  be  taken  as  a  factor.  Thus 
S2  expresses  the  square  of  5;  73  the  cube  of  7;  86  the  fifth  power  of  8;  etc. 

315.  The  root  of  a  power  is  one  of  the  number  of  equal  fac- 
tors of  which  that  power  is  the  product. 


POWERS   AND   ROOTS  275 

NOTE  1.  The  square  root  of  a  power  is  one  of  its  two  equal  factors;  the 
cube  root  is  one  of  its  three  equal  factors;  the  fourth  root  is  one  of  its  four  equal 
factors;  etc. 

NOTE  2.  The  radical  sign  V~  is  usually  written  at  the  left  of  a  power  to 
indicate  that  a  root  of  it  is  to  be  taken;  and  a  small  figure  called  an  index  is 
written  within  the  angle  of  the  radical  sign  to  show  what  root  is  to  be  taken. 
When  no  index  is  written,  the  index  2  is  understood.  Thus,  VT^  indicates 
the  square  root  of  16;  -^512  the  cube  root  of  512;  v'ST  the  fourth  root  of  81. 

316.  Principles.     1.   //,    commencing    at     the     right    of    any 
given  power  of  an  integral  number,  it  is  divided  by  a  separatrix  into 
sections  of  as  many  figures  each  as  there  are  units  in  its  exponent, 
the  number  of  such  sections,  counting  a  possibly  incomplete  section 
at  the  extreme  left  as  another  section,  will  denote  the  number  of 
orders  in  its  root:  and  the  enumeration  of  those  sections  will  indicate 
the  relative  value  of  their  corresponding  root  orders. 

2.  (a)  The  highest  or  left-hand  section  contains  the  given  power 
of  the  highest  root  order:  (b)  any  excess  of  the  highest  section  over 
the  given  power  of  the  highest  root  order,  written  at  the  left  of  the  next 
following  section,  will  approximately  express  the  value  of  the  given 
power  of  the  second  highest  root  order  and  its  involved  relations  with 
the  highest  root  order:  (c)  any  excess  of  the  highest  two  sections  over 
the  given  power  of  the  highest  two  root  orders,  written  at  the  left  of 
the  next  following  section,  will  approximately  express  the  value  of  the 
given  power  of  the  third  highest  root  order  and  its  involved  relations 
with  the  highest  two  root  orders;  etc. 

ILLUSTRATIVE  EXAMPLES 

317.  What  is  the  square  root  of  4624? 

SOLUTION 

46|24(6  tens  702  =  4900 

62  =  36      g  602  =  3600 


130)1024(y  units  10)1300 

910  Average  square  per  unit  =  130 

Tl4 
V4624  =  6  tens  and  8  units,  or  68. 

EXPLANATION,  (a)  Commence  at  the  right  of  the  given  power  and 
divide  it  by  a  separatrix  into  sections  of  two  figures  each,  thus  finding  that  the 
required  root  contains  two  orders,  tens  and  units  (Prin.  1,  316). 


276  APPENDIX 

(6)  The  greatest  number  the  square  of  which  is  contained  in  the  left-hand 
or  tens'  section  (46)  is  6  tens.  Write  6  tens  as  the  highest  order  of  the  re- 
quired square  root,  subtract  its  square  (62  =  36)  from  its  corresponding  sec- 
tion (46),  to  the  remainder  (10)  bring  down  the  next  following  section  (24)  and 
the  result  (1024)  will  approximately  express  the  square  of  the  next  following 
or  units'  order  of  the  required  root  +  its  involved  relations  with  the  tens' 
order  of  the  root  (6,  Prin.  2,  316). 

(c)  The  required  square  root  is  thus  found  to  be  more  than  6  tens  (or 
60  units  if  expressed  in  terms  of  the  next  lower  root  order),  and  less  than  7 
tens  (or  70  units  if  similarly  expressed);    and  if  the  difference  between  the 
square  of  60  units  (602  =  3600)  and  the  square  of  70  units  (702  =  4900),  or 
1300,  be  divided  by  the  difference  between  their  respective  roots  (70  —  60), 
or  10,  the  quotient  (130)  will  be  the  average  square  of  the  ten  consecutive 
numbers  between  60  and  70;  and  as  the  difference  is  so  slight  as  to  be  prac- 
tically negligible,  it  (130)  can  be  accepted  tentatively  as  the  average  square 
of  as  many  of  these  consecutive  numbers  as  may  be  included  in  the  next 
following  or  units'  order  of  the  required  root. 

(d)  Divide  the  square  of  the  units'  order  of  the  required  root  and  its 
involved  relations  as  found  in  b  (1024)  by  the  assumed  average  square  per  unit 
and  its  involved  relations  as  found  in  c  (130)  to  find  tentatively  the  number  of 
units  in  the  units'  order  (7  units  and  a  significant  remainder,  practically  8 
units).     Hence,  6  tens  and  8  units,  or  68,  is  the  required  square  root. 

NOTE.  In  the  above  solution,  the  average  square  of  the  ten  consecutive 
squares  between  602  and  702  (130)  is  slightly  greater  than  the  average  square 
of  the  eight  consecutive  squares  between  602  (3600)  and  the  square  of  the  true 
root  (682  =  4624)  which  is  128  (4624  -  3600  -^  8).  Hence,  the  final  re- 
mainder (114)  from  dividing  by  the  assumed  average  square  (130)  lacks 
(130-128)  X  8,  or  16  of  containing  the  tentative  divisor  (130)  exactly  8  times. 

318.   What  is  the  cube  root  of  105823817? 

SOLUTION 

105|823|817(4  hundreds  503  =  125000 

43  =  64  7  403  =  64000 

6100)41  823(J5  tens  10)61000 

36600  Average  cube  per  unit  =  6100 
5223 

105823  4803  =  110592000 

473  =  103  823        3  4703  =  103823000 

676900)2  000  817$  units  10)6769000 

1  353  800  Average  cube  per  unit  =  676900 

647  017 
Hence,  s/105823817  =  4  hundreds,  7  tens,  3  units,  or  473. 


POWERS   AND   ROOTS  277 

EXPLANATION,  (a)  Commence  at  the  right  of  the  given  cube  and  divide 
it  by  a  separatrix  into  sections  of  three  figures  each,  thus  finding  that  the 
required  root  contains  three  orders,  hundreds,  tens,  and  units  (Prin.  1,  316). 

(6)  The  greatest  number  the  cube  of  which  is  contained  in  the  left-hand 
or  hundreds'  section  (105)  is  4  hundred.  Write  it  (4  hundred)  as  the  highest 
order  of  the  required  root  (a,  Prin.  2,  316),  subtract  its  cube  (4s  =  64)  from  its 
corresponding  section  (105),  to  the  remainder  (41)  bring  down  the  next  fol- 
lowing section  (823),  and  the  result  (41823)  will  approximately  express  the 
cube  of  the  next  following  or  tens'  order  of  the  required  root  +  its  involved 
relations  with  the  hundreds'  order  of  the  root  (b,  Prin.  2,  316). 

(c)  It  is  now  found  that  the  required  cube  root  is  greater  than"  4  hundred 
(or  40  tens  if  expressed  in  terms  of  the  next  lower  root  order)  and  less  than  5 
hundred  (or  50  tens  if  similarly  expressed) .     If  therefore  the  difference  between 
the  cube  of  40  tens  (40*  =  64000)  and  the  cube  of  50  tens  (50»  =  125000),  or 
61000,  be  divided  by  the  difference  between  their  respective  roots  (50  -  40), 
or  10,  the  quotient  (6100)  will  be  the  average  cube  of  the  ten  consecutive 
numbers  between  40  and  50;  and  as  the  difference  is  so  slight  as  to  be  prac- 
tically negligible,  this  result  (6100)  can  be  accepted  tentatively  as  the  average 
cube  of  as  many  of  these  consecutive  numbers  as  may  be  included  in  the  next 
following  or  tens'  order  of  the  required  root. 

(d)  Divide  the  cube  of  the  tens'  order  of  the  required  root  and  its  involved 
relations,  as  found  in  b  (41823)  by  the  assumed  average  cube  per  unit  and  its 
involved  relations  as  found  in  c  (6100)  to  find  tentatively  the  number  of  units 
in  the  tens'  order  (6  tens  and  a  significant  remainder,  practically  7  tens). 

(e)  The  required  root  is  now  found  to  be  greater  than  47  tens  (or  470  units 
if  expressed  in  terms  of  the  next  lower  root  order)  and  less  than  48  tens  (or 
480  units  if  similarly  expressed).     If  therefore  the  difference  between  470» 
(103823000)  and  4801  (110592000),  or  6769000,  be  divided  by  the  difference 
between  their  respective  roots  (480  —  470),  or  10,  the  quotient  (676900)  will 
be  the  average  cube  of  the  ten  consecutive  numbers  between  470  and  480; 
which  may  also  be  accepted  tentatively  as  the  average  cube  of  as  many  of 
these  consecutive  numbers  as  are  included  in  the  next  following  or  units'  order 
of  the  required  root. 

(/)  Subtract  the  cube  of  the  already  obtained  root  orders  (47*  =  103823) 
from  their  corresponding  sections  (105823),  to  the  remainder  (2000)  bring  down 
the  next  following  section  (817),  and  the  result  (2000817)  will  approximately 
express  the  cube  of  the  next  following  or  units'  order  of  the  required  root  +  its 
involved  relations  with  the  preceding  root  orders  (c,  Prin.  2,  316). 

(g)  Divide  the  cube  of  the  units'  order  of  the  required  root  and  its  involved 
relations  as  obtained  in  /  (2000817)  by  the  assumed  average  cube  per  unit  in 
the  units'  order  and  its  involved  relations  as  obtained  in  e  (676900)  to  find 
tentatively  the  number  of  units  in  the  units'  order  (2  units  and  a  significant 
remainder,  or  3  units).  Hence,  4  hundreds,  7  tens,  and  3  units,  or  473,  is  the 
required  cube  root. 


278  APPENDIX 

319.  In  the  following  solutions,  the  average  power  per  unit 
is  more  briefly  obtained  by  finding  the  difference  between  the 
given  power  of  the  root  orders  already  found  and  the  given  power 
of  those  root  orders  increased  by  1  without  expressing  them  in  terms 
of  the  next  lower  root  order,  and  annexing  to  this  difference  as  many 
ciphers  lacking  1  as  there  are  units  hi  the  exponent  of  the  given 
power.  Much  unnecessary  labor  may  also  be  avoided  in  the  suc- 
cessive divisions  to  find  the  next  following  root  order  by  using  only 
the  highest  two  orders  of  the  successive  divisors,  and  by  omitting 
from  the  right  of  their  corresponding  dividends  as  many  figures 
as  have  been  rejected  from  the  right  of  their  respective  divisors. 
The  processes  being  identical  for  all  roots,  no  further  explanations 
are  considered  necessary.  If,  however,  the  learner  should  fail 
to  understand  any  step  hi  the  following  solutions,  he  can  refer 
to  the  corresponding  step  in  the  preceding  explanations  for  the 
required  information. 

ILLUSTRATIVE  EXAMPLES 
1.   What  is  the  fifth  root  of  3796375994368? 

SOLUTION 

379|63759|94368(3  hundred 
35=243  2 

78|10000)136|63759(;  tens  45  =  1024 

78  35  =    243 


58  Average  power  per  unit  =    7810000 

37963759 

32 5  =  33554432  g  33 5  =  39135393 

55|809610000)440|932794368(/f  units  325  =  33554432 


Average  power  per  unit  =    55809610000 
50 
N/3796375994368  =  3  hundred,  2  tens,  8  units,  or  328. 

2.   What  is  the  fourth  root  of  552114385936? 


POWERS   AND    ROOTS  279 

SOLUTION 

5521|1438|5936(8  hundred 
84  =  4096          6 

24|65000)142|51438(P  tens  94  =  6561 

123  84  =  4096 


19  Average  power  per  unit  =  2465000 

55211438 

86 4  =  54700816  87 4  =  57289761 

25|88945000)51 106225936(2  units  864  =  54700816 

51  Average  power  per  unit  =    2588945000 


<</5521 14385936  =  8  hundred,  6  tens,  2  units,  or  862. 
3.   What  is  the  square  root  of  .02795584? 

SOLUTION 

.02|79|55|84(1  tenth 
.I2  =.01      6 

30)1 79  (£  hundredths  22  =  4 

150  I2  =  1_ 

29  Average  per  unit  =  30 

.0279  17 2  =  289 

.162  =  .0256  162  =  256 


330)2355(7  thousandths  Average  per  unit  =    330 

2310 

45,  insignificant  remainder 

.027955 
.1672  =  .027889      2 

3350)       6684(;  ten-thousandths  168 2  =  28224 

3350  167 2  =  27889 


Average  per  unit  =      3350 

V.02795584  =  1  tenth,  6  hundredth,  7  thousandth,  2  ten-thou- 
sandth, or  .1672. 


280  APPENDIX 

RULE,  (a)  Commence  at  the  right  of  the  given  power,  if  an 
integer,  and  separate  it  into  sections  of  as  many  figures  each  as  there 
are  units  in  the  exponent  of  that  power.  The  number  of  sections 
thus  found,  counting  a  possibly  incomplete  section  at  the  extreme 
left  as  another  section,  will  denote  the  number  of  orders  in  the  required 
root,  and  the  enumeration  of  those  sections  will  indicate  the  relative 
value  of  their  corresponding  root  orders. 

(b)  Find  the  greatest  number  the  given  power  of  which  is  con- 
tained in  the  left-hand  section,  and  write  it  as  the  highest  order  of 
the  required  root. 

(c)  Subtract  the  given  power  of  the  obtained  highest  root  order 
from  the  left-hand  section,  and  to  the  remainder  bring  down  the  next 
following  section,  to  find  the  dividend  of  the  next  following  root  order. 

(d)  Find  the  difference  between  the  given  power  of  the  obtained 
root  order  and  the  given  power  of  that  order  increased  by  1,  and  to 
this  difference  annex  as  many  ciphers  less  1  as  there  are  units  in 
the  exponent  of  the  given  power,  to  find  tentatively  the  average  given 
power  per  unit  of  the  next  following  root  order. 

(e)  Divide  the  obtained  dividend  of  the  next  following  root  order 
(c)  by  the  average  given  power  per  unit  of  the  next  following  root 
order  (d)  to  find  tentatively  the  number  of  units  in  the  next  following 
root  order,  increasing  the  quotient  by  1  if  the  remainder  be  signifi- 
cantly great  as  compared  with  the  divisor. 

(f)  Thus  continue  to  subtract  the  given  power  of  the  obtained 
root  orders  from  their  corresponding  sections,  to  the  remainder  bring 
down  the  next  following  section,  and  divide  the  result  by  the  average 
power  per  unit  of  the  next  following  root  order  to  find  tentatively  the 
number  of  units  in  that  root  order,  until  all  the  root  orders  have  been 
successively  obtained. 

NOTE  1.  In  applying  e  of  the  above  Rule,  if  the  quotient  from  dividing 
the  dividend  of  the  next  following  root  order  by  the  divisor  of  that  root  order, 
should  be  1  greater  or  1  less  than  the  true  root  order,  as  is  sometimes  the  case, 
it  will  incidentally  be  made  evident  in  performing  the  next  following  process. 
Thus,  if  the  given  power  of  the  obtained  root  orders  should  prove  to  be  greater 
than  the  sections  from  which  they  were  evolved,  it  would  show  that  the  last 
obtained  root  order  should  be  diminished  by  1 ;  or  if  the  given  power  of  the 
obtained  root  orders  plus  1  should  prove  to  be  equal  to  or  less  than  the  sections 
from  which  the  obtained  root  orders  were  evolved,  it  would  show  that  the  last 
obtained  root  order  should  be  increased  by  1. 


POWERS   AND    ROOTS 


281 


NOTE  2.  The  roots  of  pure  decimals  are  obtained  in  the  same  manner 
as  those  of  integers,  except  that  decimals  should  be  separated  into  sections  by 
commencing  at  the  left  (at  the  decimal  point).  Mixed  decimals  are  divided 
into  sections  by  commencing  at  the  decimal  point  and  separating  the  integral 
part  into  sections  from  the  right,  and  the  decimal  part  into  sections  from  the 
left. 

NOTE  3.  The  root  of  a  common  fraction  is  obtained  by  separately  finding 
the  root  of  its  numerator  and  of  its  denominator.  If  the  numerator  and 
denominator  are  not  perfect  powers,  first  reduce  to  a  decimal,  and  apply 
Note  2. 

NOTE  4.  If,  after  bringing  down  and  considering  the  last  section  of  an 
integral  number,  it  is  found  to  be  an  imperfect  power,  continue  the  process 
by  successively  annexing  a  section  of  ciphers  as  many  times  as  there  may  be 
decimal  places  required  in  the  root.  If  preferred,  the  obtained  decimal  part 
of  the  root  can  then  be  approximated  to  a  common  fraction  of  any  required 
denominator  by  125. 


EXAMPLES  FOR  PRACTISE 

Applying  314,  find  the  indicated  power  of 

1.  92  5.   83  9.   54        13.    (f)3 

2.  282          6.    153          10.    135      14.    (j)2 


17.  (.6)5 

18.  (.27) 4 


3.  162 2 

4.  345 2 


7.  246 3 

8.  528 3 


11.  6' 

12.  86 


15.  (2TV)4     19.    (1.28)3 

16.  (5i)3       20.    (13.72) 


Find  the  indicated  root  of 

21.  V5476 

22.  V132496 

23.  V19321 

24.  V609961 

25.  V169744 

26.  V82864609 

27.  V74701449 


29.  V2394829969  37. 

30.  V391Q50115"6  38. 

31.  ^804357    39. 

32.  \X425259008'  40. 

33.  \X14886936   41. 

34.  ^997002999  42. 

35.  \V177504328  43. 


28.  V14561856  36.  ^575930368  44. 


^9340607016 

\X823183846848 

\XlQ03Q03001 

\V41962796875 

V 28398241 

^169879162896 

^3707398432 

-yj/ 1099 144686 11443 


45.  Allowing  2150.42  cu.  in.  to  a  bushel,  what  must  be  the 
dimensions  of  a  cubic  bin  which  will  hold  75  bushels  of  wheat, 
carrying  the  dimensions  to  two  decimal  places? 

46.  Allowing  231  cu.  in.  to  a  gallon,  what  are  the  dimensions 
of  a  cubic  cistern  which  will  hold  30000  gallons  of  water  —  to  three 
decimal  places? 


282  APPENDIX 

47.  How  many  feet  long  is  each  side  of  a  square  field  which 
contains  5  acres  —  to  two  decimal  places? 

SUGGESTION.  Acres  should  be  reduced  to  square  units  of  the  denomina- 
tion in  which  the  linear  dimensions  are  required  to  be  expressed  [146]. 

48.  If  each  edge  of  a  cube  is  23  inches  long,  what  are  its  con- 
tents? 

49.  If  each  side  of  a  square  lot  is  135  ft.  6  in.  long,  what  is 
its  area? 

60.  Allowing  34.844  cu.  ft.  to  a  ton,  what  must  be  the  dimen- 
sions of  a  cubic  bin  which  will  hold  18  tons  of  coal  —  to  two 
decimal  places? 

51.  A  rectangular  lawn  is  three-fourths  as  wide  as  it  is  long. 
If  its  area  is  768  sq.  yd.,  what  are  its  dimensions? 

SUGGESTION.  In  Ex.  51,  divide  the  given  area  by  f  to  find  the  area  of  a 
square  in  which  the  width  will  exactly  equal  the  length,  extract  the  square 
root  of  the  result  to  find  the  length,  etc. 

52.  A  rectangular  room  is  three  times  as  long  as  it  is  wide. 
What  are  its  dimensions  if  its  area  is  5547  sq.  ft.? 

53.  A  rectangular  platform  is  two-thirds  as  long  as  it  is  wide. 
If  its  area  is  1536  sq.  ft.,  what  are  its  dimensions? 

54.  A  rectangular  reservoir  which  is  six  times  as  long  as  it 
is  high  and  wide  has  a  capacity  of  3951018  cu.  ft.  of  water.     What 
are  its  dimensions? 

SUGGESTION.  In  Ex.  54,  divide  the  given  capacity  by  6  to  find  the  capacity 
of  a  cubic  reservoir  in  which  the  length  is  exactly  equal  to  the  height  and  the 
width,  extract  the  cube  root  of  the  result  to  find  the  height  and  the  width,  etc. 

55.  An  excavation  which  is  four  times  as  long  as  it  is  wide 
and  deep  required  the  removal  of  740772  cu.  ft.  of  earth.     What 
were  its  dimensions? 

56.  What  are  the  dimensions  of  a  rectangular  volume  the 
width  of  which  is  f  as  much  as  its  length  and  its  height,  if  it  con- 
tains 12288  cu.  in.? 


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(1.)  2957 
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(1.)  4694 

(6.)  49193 

(9.)  53959942 

(13.)  458891 

(17.)  68  yr. 


(1.)  17471 
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(2.)  11601 
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Article  20  (Page  9) 

(2.)  2700  (3.)  3130 

(7.)  3629  (8.)  7989 


(4.)  23845 


(4.)  3481  (5.)  14112 


Article  23  (Pages  10-11) 

(2.)  3314        (3.)  55532 

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(15.)  $18214 


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(8.)  5668586134672 

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Article  29  (Page  15) 


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284 


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Article  44  (Page  26) 

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$12750  (33.)  5625  mi.  (34.)  9814  votes  (36.)  $532.96  (36.)  1360 
mi.  (37.)  19  mi.  (38.)  $25  (39.)  40248  books  (40.)  80  coins 

Article  58  (Page  41) 

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2,  2,  3,  3,  5,  7,  7     (12.)  2,  2,  2,  2,  2,  3,  3,  3,  3,  11     (13.)  2,  2,  2,  3,  3, 

3,  7,  11    (14.)  2,  2,  2,  3,  3,  5,  5,  5,  5    (16.)  2,  2,  3,  3,  5,  7,  13 


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Article  63  (Page  42) 

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Article  67  (Page  45) 


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Article  83  (Page  50) 
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Article  84  (Page  51) 

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Article  86  (Page  52) 
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Article  87  (Page  53) 

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Article  88  (Page  53) 

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Article  91  (Page  55) 

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Article  92  (Pages  56-57) 

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Article  94  (Page  58) 

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287 


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(6.)  378 
(11.)  31 


Article  106  (Page  69) 


(3.) 
(8.) 
(13.)  37 


(4.) 

(9.)  41H"ft" 

(14.)  151A 


Article  108  (Pages  70-71) 


(1.)  $M       (2.)  6  yd.       (3.)  2  pk.       (4.)  8  Ib.       (6.) 
(7.)  M  hr.          (8.)  3  da.          (9.)  13  da.          (10.)  95  da. 
(12.)  $8TA*       (130  23T%  yd.       (14.)  A       (160  7J  gal. 


(10.)  6 
(16.)  1 


hr.       (6.)  A  A. 

(11.)  7^  mi. 

(16.)  108A  yd. 


Article  110  (Pages  74-75) 


(1.)  $6300 
(7.)  H       (8  ) 


(2.)  60  Ib.        (3.)  A 
0.)  171  sheep 


(4.)  $5628        (6.)  $2316        (6.)  f 
(10.)  *       (11.)  $7800       (12.)  $18275 


288 


ANSWERS 


Article  111  (Pages  75-77) 

(1.)  2990H  (20  26 j\  acres  (3.)  $21.83ff  (4.)  J»»-  (6.)  21HH 
mi.  (6.)  $1.831  (7.)  $12710A  (8.)  1900  mi.  (9.)  19^  hr.  (10.) 
$3943.78i  (11.)  TV  da.  (12.)  ff  (13.)  grammar  $.69;  speller  $.23 
(16.)  2-Hda.  (16.)  2Hda.  (17.)  30  da.  (18.)  81U  (19.)  U 

(20.)  884£  (21.)  212TV  Ib.  (22.)  400  Ib.  (23.)  10  min.  (24.)  $450 
(25.)  f  (26.)  $13450;  $10760  (27.)  $12800  (28.)  568i  (29.)  454^ 
(30.)  f  (31.)  A  $4280;  B  $3210;  C  $2568  (32.)  $2125 

Article  123  (Page  83) 

(1.)  .5  (2.)  .75  (3.)  .625  (4.)  .25  (5)  .875  (6.)  .375  (7.) 
.3125  (8.)  .28125  (9.)  .125  (10.)  .140625  (11.)  .6875  (12.)  .8 
(13.)  .36  (14.)  .136  (16.)  .203125  (16.)  18.1875  (17.)  465.21875 
(18.)  67.49609375  (19.)  .71875  (20.)  .328125  (21.)  .9375  (22.) 
.34375  (23.)  .109375  (24.)  .8125  (26.)  .546875  (26.)  .03125  (27.) 
58.4375  (28.)  63.53125  (29.)  37.046875  (30.)  78.0625  (31.) 
168.44-  (32.)  34.06-  (33.)  72.72-  (34.)  91.688-  (36.)  26.094- 
(36.)  49.016-  (37.)  23.4688-  (38.)  71.7656+  (39.)  37.8125  (40.) 
8.26563-  (41.)  7.34867-  (42.)  9.52722+  (43.)  1.276431+  (44.) 
3.296429-  (45.)  2.834538- 


(1.)  f 


)•)  16T3T 


(2.)  f 

(9.)  A 

(15.)  8£ 


Article  124  (Page  84) 

(3.)  rir         (4.)  |         (5.)  -;ff         (6.) 
(10.)  H       (ID  if       (12.)  riff       (13.) 
(16.)  2Tfo          (17.)  19§          (18.)  6* 


(7.)  H 

(14.) 

(19.)  9* 


(1.) 
(8.) 


(2.)  ) 
(9.)  I 


Article  125  (Page  85) 
(3.)  f        (4.)  f        (5.)  | 

Article  126  (Pages  86-87) 

(1.)  923.6129    (2.)  80.9235       (3.)  816.19108 

(6.)  5.65395     (6.)  626.109375     (7.)  76.250375 

(9.)  123.079-   (10.)  12.5576+     (11.)  20.676- 

(13.)  136137.1064684    (14.)  9341.9448348 


(6.)  i     (7.) 


(4.)  656.82651 

(8.)  92.603845 

(12.)  1128.037265 


Article  127  (Page  87) 


(1.)  46.786 

(6.)  75.383 

(9.)  14.175 

(13.)  6.7535625 

(17.)  231.9944 


(1.)  22221.54 

(6.)  .54528 

(9.)  4.774225 

(13.)  6.15470625 

(17.)  2.7500^ 

(21.)  7.183 


(2.)  580.8655 

(6.)  24.747 

(10.)  .53133 

(14.)  28.103  + 

(18.)  .77427 


(3.)  38.4322 

(7.)  257.1286 

(11.)  26.0675 

(16.)  67.2320- 


Article  128  (Pages  88-89) 


(2.)  590.7568 

(6.)  8.58426 

(10.)  20.660234375 

(14.)  4.02636 

(18.)  .3847^ 


(3.)  33.20697 
(7.)  .003973 
(11.)  24.768 
(15.)  10.138^ 
(19.)  71.83 


(4.)  34.858 

(8.)  67.6871 

(12.)  142.50125 

(16.)  5.83  + 


(4.)  .070924 

(8.)  39.845 
(12.)  15.181809375 
(16.)  1.801  Hi 
(20.)  620 


ANSWERS 


289 


Article  129  (Page  90) 


(1.)  5.447 

(5.)  46250 

(9.)  190000 

(13.)  4.0625 

(17.)  1.1353- 

(21.)  12.782- 

(25.)  90.0671  + 

(29.)  .052384 

(33.)  .0969- 


(1.)  $18.80- 
(6.)  $1394.54 
(9.)  $14.86- 


(2.)  2.8 
(6.)  .0682 

(10.)  .0004 

(14.) 

(18.) 

(22.) 

(26.) 

(30.) 

(34.) 


78.36 

12.82- 

.526  + 

.55 

.034f 

.1091- 


(3.)  .572 

(7.)  .006 
(11.)  27000 
(15.)  4.36 
(19.)  .051- 
(23.)  946.92  + 
(27.)  2.8345 
(31.)  .067- 


Article  130  (Page  92) 


(2.)  $13.33  + 
(6.)  $4.36+ 
(10.)  $15.11  + 


(3.)  $41.17+ 
(7.)  $5.40- 


(4.)  2.5837 
(8.)  .0000032 
(12.)  125 
(16.)  726.8 
(20.)  22.83- 
(24.)  75.035  + 
(28.)  416.78 
(32.)  .017833  + 


(4.)  $151.30+ 
(8.)  $35.67+ 


Article  131  (Pages  92-94) 


(1.)  46.2394842  (2.)  2866.44921875  bu.  (3.)  361.39525  (4.)  H 
(5.)  $240.885+  (6.)  73  (7.)  .75  yd.  (8.)  637.5  bu.  (9.)  .375 
(10.)  432  brls.  (11.)  $9600  (12.)  .74  (13.)  $1560  (14.)  wife 
$13566;  daughter  $8996.40;  son  $5997.60  (15.)  house  $2000;  lot  $1500 
(16.)  $16400  (17.)  $82.40  (18.)  $34  (19.)  $468  (20.)  5400  inhabi- 
tants (21.)  A  .46;  B  .54  (22.)  $710.10  (23.)  .28  (24.)  $95  (25.) 
$396 

Article  156  (Page  100) 

(1.)  9583  d.    (2.)  147600  Ib.    (3.)  1082  pt.    (4.)  19080  m. 
113271120  sec.    (6.)  103944  gr.    (7.)  47234  d.    (8.)  5628  in. 
3248  cu.  ft.   (10.)  26091396  sq.  in.   (11.)  1714356  in.   (12.)  211456  oz. 
(13.)  2221200  sec.    (14.)  1139  qt.    (15.)  2108  gi.    (16.)  460858572 
sq.  in.   (17.)  1188864  cu.  in.   (18.)  $6.72   (19.)  $412.65   (20.)  $14.80 


a) 


Article  157  (Page  101) 

(1.)  96  bu.  1  pk.  6  qt.  1  pt.  (2.)  41  da.  6  hr.  54  min.  42  sec.  (3.)  4 
mi.  101  rd.  2  ft.  3  in.  (4.)  18  A.  158  sq.  rd.  28  sq.  yd.  8  sq.  ft.  72  sq.  in. 
(5.)  £24  8s.  lid.  (6.)  26  T.  13  cwt.  60  Ib.  4  oz.  (7.)  11  rm.  15  sh. 
(8.)  61  cd.  1  cd.  ft.  11  cu.  ft.  (9.)  9  Ib.  9  oz.  6  pwt.  (10.)  437  cd. 
(11.)  £32  3d.  (12.)  $53  86£  7  m.  (13.)  196  gal.  1  qt.  1  pt.  3  gi. 
(14.)  33  da.  3  hr.  24  min.  12  sec.  (15.)  9  mi.  290  rd.  5  yd.  1  ft.  1  in. 
(16.)  233  bu.  (17.)  165  mi.  311  rd.  4  yd.  1  ft.  6  in.  (18.)  19  cu.  yd. 
17  cu.  ft.  1505  cu.  in.  (19.)  £74  6s.  (20.)  43  gal.  2  gi. 


Article  158  (Page  102) 


(1.)  17s.  6d. 


(2.)  273  da.  18  hr.    (3.)  1  pt.  2  gi. 
136  da.  21  hr.    (6.)  142  rd.  1  yd.  8  in. 
3  cwt.  75  Ib.   (9.)  13s.  9d.   (10.)  2  ft.  9  in. 


(4.)  2  pk.  2  qt. 
(7.)  2  ft.  6  in. 


Article  159  (Page  103) 


1.)  1  pt.  2  gi. 


(4.)  3  qt.  3  gi. 
(8.)  10s.  5.7d. 


(2.)  13  da.  16  hr.  30  min.      (3.)  2  pk.  1  qt.  1  pt. 
(5.)  17  cwt.  52  Ib.    (6.)  4  oz.  10  pwt.    (7.)  110  rd. 
(9.)  3  pk.  5  qt.  1  pt. 


290  ANSWERS 

Article  160  (Pages  103-104) 

(!.)£!  (2.)  $|  (3.)  f  gal.  (4.)  f  bu.  (6.)  *  da.  (6.)  .875  gal. 
(7.)  £.625  (8.)  -5yd.  (9.)  .4844- bu.  (10.)  .222- da. 

Article  161  (Page  104) 

(1.)  44  da.  10  hr.  3  min.  18  sec.  (2.)  106  T.  56  Ib.  13  oz.         (3.)  £66 

3s.  lid.  1  far.  (4.)  16  bu.  2  pk.  4  qt.  1  pt.  (6.)  £199  3s.  2d.  2  far. 

(6.)  58  da.  5  hr.  49  min.  29  sec.       (7.)  97  cu.  yd.  2  cu.  ft.  1639  cu.  in.      (8.) 
2  da.  10  hr.  56  min.       (9.)  3  cd.  ft.  14  cu.  ft. 

Article  162  (Page  105) 

(1.)  6  bu.  3  pk.  2  qt.  1  pt.  (2.)  25  da.  17  hr.  15  min.  (3.)  16  cd.  70 
cu.  ft.  (4.)  90  gal.  3  qt.  2  gi.  (6.)  £216  15s.  3d.  (6.)  7  A.  142 
sq.  rd.  12  sq.  yd.  21  sq.  ft.  36  sq.  in.  (7.)  11  gal.  1  qt.  1  pt.  (8.)  £187 
9s.  9d. 

Article  163  (Page  106) 

(1.)  19  yr.  8  mo.  17  da.  (2.)  25  yr.  6  mo.  6  da.  (3.)  36  yr.  7  mo. 
10  da.  (4.)  15  yr.  7  mo.  21  da.  (6.)  64  yr.  2  mo.  8  da.  (6.)  7  yr. 
10  mo.  12  da.  (7.)  13  yr.  10  mo.  (8.)  33  yr.  8  mo.  (9.)  18  yr.  (10.) 
25  yr.  1  da. 

Article  164  (Page  107) 

(1.)  300  da.  (2.)  228  da.  (3.)  270  da.  (4.)  232  da.  (5.)  225  da. 
(6.)  79  da.  (7.)  269  da.  (8.)  259  da.  (9.)  297  da.  (10.)  9  da. 
(11.)  191  da.  (12.)  290  da. 

Article  165  (Page  108) 

(1.)  45  gal.  (2.)  £  169  3s.  lid.  (  3.)  151  da.  23  hr.  15  min.  (4.) 
280  ft>.  3  oz.  12  pwt.  (5.)  643  bu.  3  pk.  4  qt.  (6.)  £26561  14s.  7d. 
(7.)  1252  mi.  45  rd.  5  ft.  6  in.  (8.)  4560  yr.  5  da.  (9.)  11587  bu.  2  pk. 
7  qt.  1  pt.  (10.)  25040  T.  2  cwt.  52  Ib.  (11.)  1615  da.  22  hr.  13  min. 
(12.)  2725  gal.  2  qt.  (13.)  13  mi.  64  rd. 

Article  166  (Page  109) 

(1.)  7  gal.  3  qt.  1  pt.  (2.)  9  yd.  2  ft.  7  in.  (3.)  £32  5s.  6d.  (4.) 
17  cd.  34  cu.  ft.  (5.)  23  bu.  3  pk.  7  qt.  (6.)  £3  7s.  8d.  (7.)  3  T. 

17  cwt.  94  Ib.         (8.)  7  gal.  3  qt.  3£f  gi.         (9.)  3  yr.  9  mo.  18  da.         (10.) 
3°  27'  35" 

Article  167  (Page  110) 

(1.)  28  times      (2.)  128  times      (3.)  59  times      (4.)  73  times      (5.)  48  times 


Article  168  (Pages^O-lll) 


(1.)  2210  bu.  3  pk.  2  qt.  (2.)  24  spoons  (3.)  $787.50  (4.)  311  da. 
(5.)  $42.43+  (6.)  43  yd.  (7.)  $11.90  (8.)  6  Ib.  9  oz.  5  pwt.  (9.) 
1  bu.  3  pk.  5  qt.  (10.)  10  sq.  rd.  25  sq.  yd.  1  sq.  ft.  126  sq.  in.  (11.) 
£738  3s.  4d.  (12.)  23  gal.  1  qt.  2  gi.  (13.)  259  mi.  125  rd.  2  yd. 
(14.)  $10.17  (16.)32brl.  (16.)  $207.88+  (17.)  $195.31-  (18.) 
$3275.76+  (19.)  $629.05+  (20.)  $173.43-  (21.)  $146.36+  (22.) 
3501.55+  fr.  (23.)  £108  11s.  5+d.  (24.)  1771.84+  rubles  (25.) 
1288.86+  pesetas. 


ANSWERS  291 

Article  180  (Page  115) 

(1.)  461348  sq.  dm.  (2.)  3468.3457  Dg.  (3.)  46817.3  dg.  (4.) 

56.78  S.  (5.)  19.653824  Ha.  (6.)  28246000  mm.  (7.)  341.7283  Dl. 
(8.)  166.52716  mi.  (9.)  147.9352  gal.  (10.)  1347.8125  bu.  (11.) 

177.15672  Dl.  (12.)  56732.48  dg.  (13.)  68.58015  M.  (14.)  290.528028 
Sq.  M.  (16.)  566.2035  M.  (16.)  11.922348  HI.  (17.)  $1028.34+ 
(18.)  3949.53-  fr. 

Article  185  (Page  117) 

(1.)  85  sq.  yd.  3  sq.  ft.  (2.)  336  sq.  yd.  (3.)  5  sq.  yd.  7  sq.  ft.  72 

sq.  in.  (4.)  24  sq.  yd.  6  sq.  ft.  (5.)  635  sq.  yd.  5  sq.  ft.  (6.)  4940 
sq.  yd.  7  sq.  ft.  72  sq.  in.  (7.)  3  A.  85  sq.  rd.  10  sq.  yd.  8  sq.  ft.  108  sq. 

in.  (8.)  1  A.  74  sq.  rd.  25  sq.  yd.  1  sq.  ft.  72  sq.  in.  (9.)  8  sq.  mi. 

19  A.  10  sq.  rd.  (10.)   114  sq.  mi.  240  A.  (11.)  9  sq.  yd.  7  sq.  ft.  18 

sq.  in.  (12.)  49  sq.  rd.  13  sq.  yd.  6  sq.  ft.  108  sq.  in.  (13.)  65.486^  sq. 
yd.  (14.)  94£  sq.  yd.  (16.)  17  A.  9  sq.  rd.  29.75  sq.  yd.  (16.)  48 

sq.  ft.  10  sq.  in.        (17.)  38  sq.  ft.        (18.)  $21        (19.)  $54 

Article  186  (Page  118) 
(1.)  5  yd.  2  ft.      (2.)  74  rd.      (3.)  28  ft.      (4.)  9  ft.        (5.)  3  rd.  1  yd.  5  in. 

Article  187  (Pages  119-121) 

(1.)  69  yd.  1  ft.  (2.)  37  yd.  1  ft.  (3.)  72  yd.  1  ft.  6  in.  (4.)  136  yd. 
(5.)  $10.80  (6.)  972  bricks  (7.)  1512  bricks  (8.)  10725  bricks  (9.)  607* 
blocks  (10.)  3569  pieces  (12.)  61  yd.  9  in.  (13.)  58  yd.  (14.)  44  yd.  2  ft. 
(16.)  93  yd.  2  ft.  2  in.  (16.)  70  yd.  1  ft.  10  in. 

Article  189  (Page  122) 

(1.)  9  bd.  ft.  (2.)  HH  bd.  ft.          (3.)  10£  bd.  ft.  (4.)  23£  bd.  ft. 

(5.)  18*  bd.  ft.          (6.)  20ff  bd.  ft.        (7.)  11H  bd.  ft.          (8.)  60  bd.  ft. 
(9.)  86  bd.  ft.          (10.)  18|  bd.  ft.        (11.)  $40.61+  (12.)  $33.54 

Article  191  (Pages  124-125) 

(1.)  672  cu.  ft.  (2.)  71  cu.  yd.  3  cu.  ft.  (3.)  10  cu.  ft.  216  cu.  in. 
(4.)  37  cu.  yd.  1  cu.  ft.  (5.)  $86.40  (6.)  $35.19+  (8.)  15  ft.  (9.)  21  ft. 
(10.)  8  ft.  (11.)  1  ft.  6  in.  (12.)  15  ft.  11+  in. 

Article  192  (Pages  126-128) 

(2.)  113.20-  bu.  (3.)  115.31+  bu.  (4.)  65.74+  bu.  (5.)  43.67+  bu. 
(6.)  2272.21-  gal.  (7.)  4578.08-  gal.  (8.)  325.82+  gal.  (9.)  22.28-  bu. 
(10.)  32315.84+  gal.  (11.)  2  ft.  5.87-  in.  (12.)  4  ft.  2.57-  in. 

(13.)  10  ft.  2.56-  in.  (14.)  128  bu.  (15.)  5  T.  896  Ib.  (16.)  4  ft.  2  in. 
(17.)  241.92  bu.  (18.)  617.14- bu.  (19.)  151.2  bu.  (21.)  2961.1008  gal. 
(22.)  564.0192  gal.  (23.)  54.12-  bu.  (24.)  9ft.  3.70-  in.  (25.)  24cu.ft. 
939.6  cu.  in.  (26.)  3694.5216  cu.  in.  (27.)  6768.2304  gal.  (28.)  125.1351 
gal.  (29.)  106.624  gal.  (30.)  80.6344  gal. 

Article  201  (Page  134) 

(1.)  $56  (2.)  $24  (3.)  182  Ib.  (4.)  714  men  (6.)  314  sheep  (6.)  730 
gal.  (7.)  $8  (8.)  6  tons  (9.)  £173  Is.  6  d.  (10.)  897  gal.  (11.)  895 


292  ANSWERS 

sheep  (12.)  192  horses  (13.)  252  da.  (14.)  476  mi.  (16.)  390  yd. 
(16.)  400  Ib.  (17.)  342  cattle  (18.)  8970  yd.  (19.)  real  est.  $12255; 
pers.  prop.  $16245  (20.)  270  pupils  (21.)  66924  inhab.  (22.)  301  acres 
(23.)  $11565 

Article  202  (Pages  136-137) 

(1.)  5  times  (2.)  .25  (3.)  33£%  (4.)  34%  (6.)  35%  (6.)  85% 
(7.)  62%  (8.)  \%  (9.)  25%  more  (10.)  20%  more  (11.)  65%  less 
(12.)  24%  less  (13.)  4%  (14.)  42%  (15.)  76%  (16.)  85%  (17.)  24% 
(18.)  A  28%;  B35%;  C  37%  (19.)  64%  (20.)  18%  (21.)  23% 

Article  203  (Pages  138-139) 

(1.)  $9  (2.)  800  gal.  (3.)  700  yd.  (4.)  560  tons  (6.)  538  gal. 
(6.)  $8496.25  (7.)  $6352  (8.)  $46764  (9.)  $386.25  (10.)  $748.75 
(11.)  $1283.75  (12.)  £781  5s.  (13.)  862.5  tons  (14.)  $327.75  (15.)  725 
acres  (16.)  $3795  (17.)  $6750  (18.)  $18000  (19.)  $45600  (20.)  $649.50 

Article  204  (Pages  140-141) 

(1.)  367  bu.  12  Ib.  (2.)  56  T.  1074  Ib.  (3.)  195  bu.  43  Ib. 

(4.)  556|  yd.  (5.)  481  bu.  4  Ib.  (6.)  125  mi.  1690  yd. 

(7.)  67  cd.  2  cd.  ft. 

Article  206  (Page  143) 

(1.)  42%          (2.)  39%        (3.)  28|%          (4.)  97%  (5.)  37f% 

(6.)  41|%         (7.)  63%        (8.)  29f%          (9.)  21.577%          (10.)  99.643% 

Article  206  (Page  144) 

(1.)  $720  (2.)  5775  natives;  2100  Germans;  1575  Irish;  1050  others 

(3.)  $1175    (4.)  $2700    (5.)  375  acres    (6.)  1200  Ib.    (7.)  $18095    (8.)  30% 

Article  207  (Page  145) 

(1.)  35%  (2.)  $1700  (3.)  $8500  (4.)  16% 

(5.)  1096518  yd.  (6.)  $45805.26         (7.)  $248856.02          (8).  $65884.50 

Article  208  (Pages  145-148) 

(1.)  $15989.19  (2.)  A  45%;  B  55%  (3.)  975  gal.  (4.)  56f%  (5.)  80% 
(6.)  A  50%  more  than  B;  B  33|%  less  than  A  (7.)  $22350  (8.)  32% 
(9.)  $5520  (10.)  650  acres  (11.)  15%  (12.)  27%  (13.)  $10890 
(14.)  $1299.80  (15.)  76%  (16.)  $14.57  (17.)  40%;  60%;  $54;  $81 
(18.)  25%;  3H%;  43f%;  $19.86;  $24.83;  $34.76  (20.)  adj.  house  $3400; 
corner  house  $4352  (21.)  A  $15720;  B  $13100;  C  $14148;  D  $10611  (22.)  A 
$84;  B  $67.20;  C$52.50;  D  $87.50;  E  $350  (23.)  $72000  (24.)  wages 
34.606%;  repairs  5.684%;  fuel  4.068%;  fixed  charges  25.855%  (25.)  4500 
bu.  (26.)  $60000 

Article  214  (Pages  149-154) 

(1.)  $14.45  (2.)  $33.08  (3.)  $77.22  (4.)  $125.70  (5.)  $57.03 
(6.)  $162.08  (7.)  $344.05  (8.)  $857.38  (9.)  $897.99  (10.)  $687.23 
(11.)  $299.88  (12.)  $382.33  (13.)  $352.46  (14.)  $52.98  (15.)  $92.44 
(16.)  $66.60  (17.)  25%  (18.)  16%  (19.)  46%  (20.)  18%  (21.)  8% 
(22.)  36%  (23.)  24%  (24.)  14%  (25.)  18%  (26.)  12*%  (27.)  20% 


ANSWERS 


293 


(28.)18f%  (29.)  16f%  (30.)  19%  (31.)  $194.60  (32.)  $468.75  (33.)  $769 
(34.)  $843.25  (36.)  $418.26  (36.)  $694.72  (37.)  $368.45  (38.)  $79.32 
(39.)  $253.60  (40.)  $62.56  (41.)  $78.50  (42.)  $625.40  (43.)  $47.26 
(44.)  $23.04  (46.)  25%  (46.)  20%  (47.)  $4.56  (48.)  $375  (49.)  $300 
(60.)  $9120  (61.)  net  loss  $200  (62.)  $1179.50  (53.)  20%  (64.)  $82.50 
(55.)  25^  (66.)  $28.13-  (67.)  30%  (68.)  18.7%  (69.)  $24000 
(61.)  37^%  (62.)  $53.25  (63.) 
15%  (65.)  $32960.88 
16^ 


exp 

13  + 


.     ,,_.,   _, .,„      (64.)  total  purchases  $61965;  gain 
DU.OO      (66.)  prod.  mat.  33^%;  prod,  labor  50%;  overhead 
(67.)  prod,  labor  47  —  %;  prod.  mat.  40+%;  overhead  exp. 
(68.)  $13    (69.)  $142.23    (70.)  $21304.12    (71.)  $8400   (72.)  $55400 


Article  220  (Page  156) 


(1.)  $241.01 
(6.)  $140.39 
(11.)  $455.38 


(2.)  $314.77 
(7.)  $414.82 
(12.)  $17.88 


(3.)  $636.87 
(8.)  $670.46 
(13.)  $551.34 


(4.)  $309.86 
(9.)  $354.12 


(5.)  $521.98 
(10.)  $222.86 


Article  221  (Page  158) 

(1.)  .72  (2.)  .64125     (3.)  .82935         (4.)  .576 

(7.)  .45448     (8.)  .59976     (9.)  .8210565    (10.)  .325 
(13.)  .2305     (14.)  .352      (15.)  .09693 


(6.) 
(11.) 


.54 
.224 


(6.)  .729 
(12.)  .6628 


(1.)  5% 
(7.)  159 


(2.)  20% 
(8.)  5% 


Article  222  (Page  159) 
(3.)  10%        (4.)  12i% 


(5.)  30%      (6.)  16f% 


Article  223  (Page  160) 


(l. 
(6. 


$48.67 
$39.57 


(2.)  $24.51 
(7.)  $76.94 


(3.)  $99.56 
(8.)  $19.90 


(4.)  $386.38 
(9.)  $12.79 


Article  231  (Page  163) 


(1.)  $5.99 
(6.)  $33.12 


(1.)  $12.26 

(6.)  $25.87 

(11.)  $42.43 


(2.) 
(7.) 


$23.60 
$35.89 


(3.)  $10.89 
(8.)  $11.52 


(4.)  $18.87 
(9.)  $240.93 


Article  232  (Page  164) 

(2.)  $55.85  (3.)  $16.18  (4.)  $4.54 

(7.)  $87.06  (8.)  $13.04  (9.)  $16.28 

(12.)  $9.57  (13.)  $55.80         (14.)  $93.42 


(6.)  $176.55 
(10.)  $96.08 


(6.)  $24.85 
(10.)  $177.44 


(5.)  $8.52 
(10.)  $37.39 


(1.) 

(6.) 
(11.) 
(16.) 
(21.) 
(26.) 
(31.) 
(36.) 
(41.) 
(46.) 

$25.93 
$5.90 
$1.34 
$.08 
$268.16 
$3 
$440.20 
$32.01 
$103.68 
$1.36 

(2.) 
(7.) 
(12.) 
(17.) 
(22.) 
(27.) 
(32.) 
(37.) 
(42.) 
(47.) 
(61.) 

Article 

$11.64 
$19.81 
$38.23 
$10 
$130.87 
$13.79 
$126.45 
$3.51 
$22.96 
$.48 
$858.30 

233 

(3.) 
(8.) 
(13.) 
(18.) 
(23.) 
(28.) 
(33.) 
(38.) 
(43.) 
(48.) 
(62.) 

(Pages  167-169) 

$30.61     (4.)  $29.68 
$42.37     (9.)  $4.33 
$2.70     (14.)  $224.99 
$41.17    (19.)  $5.79 
$339.69   (24.)  $2984.72 
$21714.51  (29.)  $10.16 
$317.77   (34.)  $33.76 
$.88      (39.)  $44.55 
$4.56     (44.)  $8.45 
$502.58   (49.)  $197.42 
$110.89   (63.)  $190.84 

(6.) 
(10.) 
(16.) 
(20.) 
(25.) 
(30.) 
(36.) 
(40.) 
(46.) 
(50.) 

$8.01 

$14.48 
$232.61 
$147.04 

$22.24 
$224.87 
$11.97 
$8.25 
$5.70 
$139.32 

294 


ANSWERS 


(1.)  $18.55 

(6.)  $7.72 
(11.)  $.71 


Article  234  (Page  170) 
(2.)  $20.31         (3.)  $15.52  (4.)  $66.98 


(7.)  $15.42 
(12.)  $472.67 


(3.)  $15.52 

(8.)  $51.44 


(9.)  $461.12 


(5.)  $10.80 
(10.)  $2.57 


(1.)  $29.56 

(6.)  $53.03 

(11.)  $232.31 

(16.)  $1022.37 


Article  235  (Page  172) 


(2.)  $10.80 
(7.)  $41.31 
(12.)  $26.23 
(17.)  $1359.58 


(3.)  $25.22 
(8.)  $40.65 
(13.)  $20.08 
(18.)  $21.93 


(4.)  $50.57 

(9.)  $820.92 

(14.)  $303.58 


(5.)  $4.09 
(10.)  $683.70 
(15.)  $684.67 


(1.)  $1.99 

(6.)  $28.25 

(11.)  $28.36 

(16.)  $211.33 


Article  236  (Page  174) 


(2.)  $6.44 
(7.)  $327 

(12.)  $17 

(17.)  $79.50 


(3.)  $1.69 

(8.)  $22.80 

(13.)  $119.88 

(18.)  $555.42 


(4.)  $3.41 

(9.)  $240 

(14.)  $20.48 

(19.)  $1638 


(6.)  $10.90 
(10.)  $17.27 
(15.)  $12.68 
(20.)  $130.05 


Article  237  (Pages  175-176) 

(1.)  $8.66  (2.)  $10.38  (3.)  $9.02  (4.)  $15.44  (5.)  $25.94  (6.)  $14.30 
(7.)  $5.29  (8.)  $24.12  (9.)  $20.72  (10.)  $117.91  (11.)  $51.57 
(12.)  $86.83  (13.)  $1811.94  (14.)  $55.64  (15.)  $475.71-  (16.)  $705.75 
(17.)  $78.36  (18.)  $526.75  (19.)  $998.68  (20.)  $846.39  (21.)  $465.13 
(22.)  £39  13s.  2.88  far.  (23.)  £492  8s.  1.05  far.  (24.)  £108  7s.  6d. 
2.09  far.  (25.)  £91  8s.  9d.  1.55  far. 


(1.)  9% 


Article  239  (Page  177) 
(2.)  5%       (3.)  5%       (4.)  4*%       (6.)  7% 


(6.) 


Article  240  (Page  178) 
(1.)  153  da.      (2.)  183  da.      (3.)  94  da.      (4.)  307  da.       (5.)  3  yr.  346  da. 

Article  241  (Page  179) 

(1.)  $582.86       (2.)  $4105.28        (3.)  $4126.40        (4.)  $9500       (5.)  $418.50 
(6.)  $7239.64         (7.)  $3875.91 


(1.) 

(6.) 


$596.52 

$2472.28 


(2.)  $1913.70 
(7.)  $951.85 


Article  242  (Page  181) 

(4.)  $3349.81 


(3.)  $741.03 
(8.)  $1589.13 


(5.)  $485.94 


Article  243  (Pages  183-185) 

(1.)  $83^85  (2.)  $165.69      (3.)  $190.72  (4.)  $767.59 

(5.)  $1053.69  (6.)  $64.89       (7.)  $48.31  (8.)  $20.40 

(9.)  $193.95  (10.)  $78.87  (11.)  $106.32  (12.)  $103.44 

(13.)  $684.78  (14.)  $1878.87  (15.)  $801.03  (16.)  $698.95 

(18.)  $2065.42  (19.)  $2906.29  (20.)  $1006.97  (21.)  $9832.58 

(22.)  $360.08  (23.)  $80.64  (24.)  $1448.19  (25.)  $593.20 


ANSWERS  295 

Article  250  (Pages  188-189) 

(1.)  $960.90;  $7.85  (2.)  $710.92;  $12.54  (3.)  $2888.88;  $76.92 
(4.)  $7725.36;  $90.99  (6.)  $511.49;  $16.41  (6.)  $838.03;  $7.09  (7.)  $647.17; 
$46.33  (8.)  $346.99;  $31.66  (9.)  $26.13  (10.)  $25.72  (11.)  $3556.10 
(12.)  Of  A;  gain  $.98  (13.)  $390.30  (14.)  $.05  (15.)  $1808.16  (16.)  $943.57 

Article  256  (Pages  191-193) 

(1.)  Aug.  15,  1911;  64  da.;  $5.30;  $591.45  (2.)  Mar.  25,  1912;  78  da.; 
$10.96;  $831.94  (3.)  Apr.  15,  1912;  37  da.;  $1.32;  $284  (4.)  Mar.  3, 
1912;  14  da.;  $4.80;  $1758.45  (5.)  Nov.  18,  1910;  26  da.;  $2.70;  $931.95 
(6.)  June  30,  1912;  79  da.;  $16.33;  $2464.79  (7.)  Dec.  23,  1911;  82  da.; 
$4.60;  $668.25  (8.)  May  8,  1911;  81  da.;  $5.47;  $481.03  (9.)  June  19, 
1912;  45  da.;  $36.74;  $3637.41  (10.)  Dec.  12,  1911;  126  da.;  $7.67;  $390.81 
(11.)  Nov.  16,  1911;  74  da.;  $8.95;  $716.35  (12.)  Nov.  30,  1911;  98  da.; 
$11.91;  $474.37  (13.)  $2546.81  (14.)  $851.29  (16.)  $721.72  (16.)  $489.83 
(17.)  $868.56  (18.)  $908.98  (19.)  $2137.26  (20.)  $466.34 

Article  257  (Page  194) 

(1.)  $656.48  (2.)  $965.97  (3.)  $1393.70  (4.)  $509.23 

(5.)  $6309.25  (6.)  $4105.37         (7.)  $698.56  (8.)  $2565.86 

(9.)  $1293.64         (10.)  $702.83         (11.)  $492.41  (12.)  $503.78 

Article  258  (Page  197) 

(1.)  $3025.70      (2.)  $1830.70      (3.)  $146.12      (4.)  $551.71      (6.)  $6612.62 
(6.)  $5194.17      (7.)  $646.42      (8.)  $358.05 

Article  259  (Pages  198-199) 

(1.)  $456.15         (2.)  $813.09         (3.)  $115.30         (4.)  $114.70 
(6.)  $223.35         (6.)  $21.54  (7.)  $100.27          (8.)  $317.96 

(9.)  $10.96 

Article  264  (Pages  201-205) 

(1.)  $18.77  (2.)  $11.35  (3.)  $4.69  (4.)  $33.72  (6.)  $23.84  (6.)  $36.17 
(7.)  $854.04  (8.)  $562.80  (9.)  $2704.65  (10.)  $8472.33  (11.)  $7313.96 
(12.)  $808.86  (13.)  $893.22  (14.)  $2393.08  (16.)  $18636.31  (16.)  $9889 
(18.)  4%  (19.)  2%  (20.)  3%  (21.)  f%  (22.)  4%  (23.)  5% 
(24.)  $685.67  (26.)  $427.60  (26.)  $916  (27.)  $17000  (28.)  $6016 
(29.)  $976.66  (30.)  $672  (31.)  $5000  (32.)  $957.99  (33.)  $650  (34.)  $500 
(35.)  $875.50  (36.)  $468.75  (37.)  $728  (38.)  $263.50  (39.)  $576 
(40.)  $852  (41.)  $9.65  (42.)  $13.39  (43.)  $10.93  (44.)  $7.05  (46.)  $422.02 
(46.)  $739.66  (47.)  157  brls.;  57£  bal.  (48.)  82  brls.;  $1.99  bal. 

(49.)297  bu.  4  Ib.  (60.)  $765.68  (61.)  $4.78  (62.)  $6097  (63.)  $4360.75 
(54.)  $379.57  (55.)  $164  brls.;  25?<  bal.  (56.)  145  brls.;  $1.84  bal. 
(67.)  694  bu.  12  Ib.  (68.)  226  brls.;  9£  bal. 

Article  272  (Pages  208-211) 

(1.)  $235y  (2.)  $217^ (3.)  $114F    (4.)  $83.13     (5.)  $909     (6.)  $2443.75      / 
(7.)  $4933.88    (8./ $3040.88    (9.)  $12894.50    (10.)  $3006.25    (11.)  $11169.25  // 
(12.)  $7463      (I/.)  $5095.50      (14.)  $11532      (15.)  $8319.38      /16.)  $1890  V  / 
(17.)  $3422.50x/  (18.)  $2734.38       (19.)  $1303.50       (20.)  4%V  (21.)  122f    S 


296 


ANSWERS 


(23.)  24J%  (24.)  19|%  (26.)  7f%  (26.)  4J%  *  (27.)  91| 
(29.)  13i%    (30.)  17|%    (31.)  5J%    (32.)  14*% 
(34.)  $1500  (36.)  $11400  (36.)  $23700  (37.)  $900  (38.)  $7100 
(40.)  $5800    (41.)  $6200    (42.)  $2300    (43.)  $4500 
(45.)  $13500  (46.)  $1200  (47.)  $4400  (48.)  $8200  (49.)  $7200 
(61.)  $11076    (62.)  $36.50   (63.)  $5194  (54.)  $940 


(22.)  161% 
(28.)  161% 
(33.)  23-1% 
(39.)  $8900 
(44.)  $3200 
(60.)  $10237.50 
(56.)  $24937.50 


Article  273  (Pages  212-214) 

(1.)  $87.50H2.)  $66    (3.)  $125   (4.)  $160    (5.)  $112.50     (6.)62f    (7.)  79 
(8.)  74£      (9.)  69|     (10.)  741      (11.)  5%      (12.)  8%      (13.)  4%      (14.)  5 
(15.)  5%       (16.)   6f%        (17.)  4%        (18.)  4|%       (19.)  6*%       (20.)  4 
(21.)  3%    (22.)  6%    (23.)  10%    (24.)  8%    (26.)  5%     (26.)  6%      (27.)  4 
(28.)  4i%    (29.)  8%     (30.)  12%    (31.)  7%     (32.)  59|    (33.)  5%    (34.)4% 
stock  yields  T\%  more  on  inv.     (35.)  $46.25     (36.)  $62.50     (37.)  $258.75 


(2.)  $631.58 

(7.)  $837.42 

(12.)  $636.96 

(17.)  $1457.91 

(22.)  $3535.35 


(1.)  $2026.27 
(6.)  $459.59 
(9.)  $274.79 


Article  278  (Pages  216-218) 

(3.)  $954  (4.)  $569.01 

(8.)  $472.12  (9.)  $622.70 

(13.)  $158.20  (15.)  $872 

(18.)  $721.91  (20.)  $754.51 

(23.)  $881.98  (24.)  $2796.39 

Article  283  (Page  220) 


(2.)  $3536.10 
(6.)  $831.54 
(10.)  $928.63 


(3.)  $4505.25 

(7.)  $901.90 

(11.)  $4025.36 


(6.)  $274.47 
(11.)  $715.42 
(16.)  $938.72 
(21.)  $1603.75 
(25.)  $3266.40 


(4.)  $198.62 

(8.)  $153.80 

(12.)  $1226.81 


Article  284  (Page  221) 


(1.)  £375  Is.  3£d.  (2.)  £153  14s.  Ifd.  (3.)  4287.45  fr.  (4.)  13384.79 
fr.  (6.)  3974.28  marks  (6.)  26249.87  marks  (7.)  1638.30  guilders 
(8.)  3054.07  guilders  (9.)  £750  10s.  6|d.  (10.)  1894.72  guilders 

Article  285  (Pages  222-223) 

(2.)  68%;  $569.50  (3.)  76%;  $1356.75          (4.)  65.28%;  A  $378.62; 

B  $832.65;  C  $571.40 

Article  286  (Page  224) 
(3.)  |%          (4.)  $65.63  (6.)  $564,75  (6.)  $319.50 


(2.)  $3500 
(7.)  $1600 


(2.)  $141.75 
(7.)  $137.60 


(2.)  $820.50 
(6.)  $4138.40 
(10.)  $3480 


(8.)  $3000;  $1500 


(9.)  $2500;  $3750;  $5000 


Article  287  (Page  225) 

(3.)  $156.75       (4.)  $184.50         (6.)  $108.75        (6.)  $405.92 
(8.)  $40.75         (9.)  $1.50  (10.)  15  mills 


Article  288  (Pages  226-227) 


(3.)  $791.10 
(7.)  $11928 
(11.)  $698 


(4.)  $2768.40 
(8.)  $51.75 
(12.)  $3598 


(5.)  $2360.75 
(9.)  $42 


ANSWERS  297 

Article  289  (Pages  229-231) 

(1.)  $57  (2.)  $282.20  (3.)  $1725  (4.)  360  Ib.  (5.)  270  sheep 
(6.)  $3980.40  (7.)  $18.18  (8.)  $16.40  (9.)  $648  (10.)  152  da.  (11.)  125  ft. 
(12.)  8  ft.  (13.)  24  men  (14.)  30  da.  (16.)  525  Ib.  (16.)  $56.25 
(17.)  A  $1500;  B  $2100  (18.)  25%  (19.)  35%  (20.)  $46.80 

Article  290  (Page  232) 
(1.)  90000  shingles    (2.)  $350    (3.)  $425    (4.)  $180    (5.)  15  da.      (6.)  $180 

Article  292  (Pages  234-235) 

(1.)  2  mo.  10  da.  (2.)  1  mo.  29  da.  (3.)  2  mo.  13  da.  (4.)  June  18,  1912 
(5.)  Jan.  3,  1912  (6.)  6  mo.  (7.)  Apr.  24,  1911  (8.)  July  14,  1910 

Article  293  (Pages  237-238) 

(1.)  May  19,  1910  (2.)  May    2,  1911  (3.)  Aug.  23,  1912 

(4.)  Aug.     4,  1911  (5.)  Nov.  24,  1910  (6.)  Apr.  25,  1912 

(7.)  Sept.  23,  1910  (8.)  Oct.  28,  1911  (9.)  May  12,  1912 

(10.)  Nov.    8,  1910  (11.)  July  30,  1911  (12.)  Dec.    5,  1912 

Article  294  (Pages  240-241) 

(1.)  July    11,  1910  (2.)  June  27,  1911  (3.)  June  23,  1912 

(4.)  Dec.  24,  1911  (5.)  Dec.  10,  1909  (6.)  Sept.  16,  1912 

(7.)  Sept.  16,  1912  (8.)  Dec.     2,  1911  (9.)  Sept.  15,  1911 

(10.)  June    4,  1912 

Article  296  (Pages  244-246) 

(1.)  Oct.    20,  1910  (2.)  Apr.  10,  1911  (3.)  May    7,  1910 

(4.)  Sept.  25,  1912  (5.)  Apr.  27,  1911  '  (6.)  Aug.    2,  1910 

(7.)  June  14,  1912  (8.)  Nov.  11,  1911  (9.)  May    3,  1910 

(10.)  Sept.  17,  1911  (11.)  June  21,  1912  (12.)  Feb.  29,  1912 

(13.)  $606.60  (14.)  Aug.  25,  1911  (16.)  Jan.  30,  1912 

Article  297  (Pages  247-248) 
(1.)  June  5,  1910    (2.)  Oct.  20,  1911    (3.)  Net  pro.  $1269.06;  May  22, 1912 

Article  299  (Pages  250-252) 

1.)  $301.47         (2.)  $353.40        (3.)  $326.90        (4.)  $521.41 

6.)  $477.74 

Article  301  (Pages  254-255) 

(1.)  $3060  gain  (2.)  $2589.88  loss  (3.)  $3786.90  loss  (4.)  $3996  gain 
(5.)  $9901.75  loss  (6.)  $11304.85  gain  (7.)  $17543.75  cap.  (8.)  $5089.77 
cap.  (9.)  $954.10  cap.  (10.)  $6899.40  insolv.  (11.)  $3012.45  insolv. 
(12.)  $5458.25  insolv.  (13.)  $5292.60  cap.  (14.)  $2959.90  insolv. 

(15.)  $11121.40  cap.  (16.)  $9868.65  cap.  (17.)  $5006.45  insolv.  (18.)  $4042.30 
cap.  (19.)  $1267  insolv.  (20.)  $3136.70  cap.  (21.)  $6508.68  (22.)  A's 
$5884;  B's  $2942  (23.)  A's  $9640;  B's  $8140;  C's  $6640 


ANSWERS 

Article  302  (Page  257) 

(1.)  A's  $1920;    B's  $1680  (2.)  A's  $2400;    B's  $3000;  C's  $3600 

(3.)  A's  $9310;  B's  $13965         (4.)  A's  $2240;  B's  $2800;  C's  $3360 

Article  303  (Pages  260-261) 

(1.)  A's  $2396.04;  B's  $1603.96  (2.)  A's  gain  $2190.08;  B's  gain  $1834.71; 
C's  gain  $2975.21;  A's  present  int.  $12190.08;  B's  present  int.  $9834.71;  C's 
present  int.  $17975.21  (3.)  A's  $26137.72;  B's  $21868.26;  C's  $16994.01 

(4.)  A's  gain  $3043.87;  A's  present  int.  $18043.87;  B's  gain  $2956.13;  B's 
present  int.  $23956.13  (6.)  A's  $12935.37;  B's  $12064.63  (6.)  A's  $19437.35; 
B's  $12562.65 

Article  304  (Pages  265-266) 

(1.)  A's  $16502.33;  B's  $21097.67  (2.)  A's  $19981.95;  B's  $24618.05 

(3.)  A's  $21837.05;  B's  $21397.60;    C's  $21065.35  (4.)  A's  $11770.85; 

B's  $20390.33;  C's  $11068.82     (6.)  A's  $16271.31;  B's  $9928.69 


Article  306  (Page  267) 


(1.)  $64 
(6.)  $139 
(11.)  $418.50 


$187      (3.)  $93 
$18.25     (8.)  $3050 
(12.)  $182.60 


(4.)  $118 
(9.)  $145.75 


(5.)  $121 
(10.)  $41.13 


(1.)  1875yd. 
(6.)  246  gal. 


Article  307  (Page  268) 
(2.)  1746  gal.    (3.)  592  Ib.    (4.)  172  bu. 


(6.)  330  yd. 


(7.)  836i  Ib.    (8.)  243f  doz. 


(1.)  $115.875 

(6.)  $352 

(11.)  $405 

(16.)  142f  yd. 

(19.)  624  bu. 


Article  309  (Page  269) 


(2.)  $64.20 
(6.)  $164.70 

(12.)  $81 

(16.)  386f  bu. 


(3.)  $197.50 
(7.)  $940.50 
13.)  $93.125 
17.)  52T\  gal. 


(4.)  $39 

(8.)  $646.30 

(14.)  463^1b. 

(18.)  425|  yd. 


Article  310  (Page  270) 


(1.)  $247 

(6.)  $382.20 

(9.)  $4056 

(16.)  362^ 

(19.)  1284i 

(24.)  61632 

(28.)  7118576 


(2.)  $448 

(6.)  $1472 

(12.)  346 

(16.)  307^ 

(20.)  2375 

(26.)  51740 

(29.)  250880 


(3.)  $324 

(7.)  $296.40 

(13.)  504| 

(17.)  232£ 

(22.)  84875 

(26.)  624232 

(30.)  67932800 


(4.)  $468.27 
(8.)  $1959.60 
(14.)  461* 
(18.)  1428| 
(23.)  42174 
(27.)  326697 


Article  311  (Pages  271-272) 

(1.)  $276.26   (2.)  $119.89   (3.)  $207.48   (4.)  $91.55  (5.)  $74.84 

(6.)  $18.12    (7.)  $66.15    (8.)  $88.51    (9.)  $27.70  (10.)  $47.87 

(11.)  $35.99   (12.)  $41.78   (13.)  $471.05  (14.)  $1253.32  (15.)  $166.74 


ANSWERS  299 

Article  312  (Page  273) 

(1.)  64  (2.)  55  (3.)  607  (4.)  5306  (5.)  529  bl. 

(6.)  6376  (7.)  6603         (8.)  85495          (9.)  884017        (10.)  9589235 

(11.)  7762236     (12.)  74366307 

Article  313  (Page  274) 

(1.)  1776  (2.)  3922  (3.)  2378  (4.)  5418  (5.)  1392  (6.)  3040 
(7.)  945  (8.)  1495  (9.)  2814  (10.)  3942  (11.)  972  (12.)  3796 

Article  319  (Pages  281-282) 

(1.)  81  (2.)  784  (3.)  26244  (4.)  119025  (5.)  512  (6.)  3375 
(7.)  14886936  (8.)  147197952  (9.)  625  (10.)  371293  (11.)  279936 
(12.)  262144  (13.)  ft  (14.)  &  (15.)  39^  (16.)  144t|  (17.)  .07776 
(18.)  .00531441  (19.)  2.097152  (20.)  188.2384  (21.)  74  (22.)  364  (23.)  139 
(24.)  781  (26.)  412  (26.)  9103  (27.)  8643  (28.)  3816  (29.)  48937 
(30.)  62534  (31.)  93  (32.)  752  (33.)  246  (34.)  999  (35.)  562  (36.)  832 
(37.)  2106  (38.)  9372  (39.)  1001  (40.)  3475  (41.)  73  (42.)  642 
(43.)  82  (44.)  643  (45.)  4  ft.  6.43+  in.  (46.)  15  ft.  10.653+  in. 

(47.)  60.16+  ft.          (48.)  7  cu.  ft.   71  cu.  in.  (49.)  67  sq.  rd.    13  sq. 

yd.  2  sq.  ft.  72  sq.  in.  (50.)  8.56+  ft.  (61.)  32  yd.  long;  24  yd.  wide 
(62.)  129  ft.  long;  43  ft.  wide  (63.)  48  ft.  wide;  32  ft.  long  (64.)  522 
ft.  long;  87  ft.  wide;  87  ft.  high  (65.)  228  ft.  long;  57  ft.  wide;  57  ft.  deep 
(66.)  32  ft.  long;  32  ft.  high;  32  ft.  wide 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

AN     INITIAL    FINE    OF    25    CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


JUM29    1933 


AUG  13    1934 


10  1937 


184J 


MAR 


3Mar'5E  DS 


REC'D 

DEC  6  -  1959 


LD  21-50rn-l,'33 


YC  24884 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


